ChemistryQ151–250 of 806 questions
Page 4 of 12 · English
151
ChemistryDifficultMCQMHT CET · 2026
Calculate the work done for the following reaction at $27^{\circ}C$: $C_2H_4(g) + H_2(g) \rightarrow C_2H_6(g)$ $(R = 8.314 \text{ J K}^{-1} \text{mol}^{-1})$ (in $\text{ J}$)
A
$2494.2$
B
$124.71$
C
$3741.3$
D
$187.07$
Solution
(A) The formula for work done in a chemical reaction is $W = -\Delta n_g RT$.
Step $1$: Calculate the change in the number of moles of gaseous species, $\Delta n_g = n_p - n_r$.
$\Delta n_g = 1 - (1 + 1) = 1 - 2 = -1 \text{ mol}$.
Step $2$: Convert temperature to Kelvin: $T = 27 + 273 = 300 \text{ K}$.
Step $3$: Substitute the values into the formula: $W = -(-1 \text{ mol}) \times (8.314 \text{ J K}^{-1} \text{mol}^{-1}) \times (300 \text{ K})$.
$W = 1 \times 8.314 \times 300 = 2494.2 \text{ J}$.
Step $1$: Calculate the change in the number of moles of gaseous species, $\Delta n_g = n_p - n_r$.
$\Delta n_g = 1 - (1 + 1) = 1 - 2 = -1 \text{ mol}$.
Step $2$: Convert temperature to Kelvin: $T = 27 + 273 = 300 \text{ K}$.
Step $3$: Substitute the values into the formula: $W = -(-1 \text{ mol}) \times (8.314 \text{ J K}^{-1} \text{mol}^{-1}) \times (300 \text{ K})$.
$W = 1 \times 8.314 \times 300 = 2494.2 \text{ J}$.
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152
ChemistryDifficultMCQMHT CET · 2026
If $2 \text{ mole}$ of an ideal gas expand isothermally and reversibly at $27^{\circ}C$ from $1 \text{ dm}^3$ to $1 \text{ m}^3$, calculate the work done. [$R = 8.314 \text{ J K}^{-1} \text{mol}^{-1}$] (in $\text{ kJ}$)
A
$-49.95$
B
$-99.90$
C
$-34.46$
D
$-68.92$
Solution
(C) Step $1$: Identify the given values.
$n = 2 \text{ mol}$, $T = 27^{\circ}C = 300 \text{ K}$, $V_1 = 1 \text{ dm}^3 = 10^{-3} \text{ m}^3$, $V_2 = 1 \text{ m}^3$, $R = 8.314 \text{ J K}^{-1} \text{mol}^{-1}$.
Step $2$: Use the formula for isothermal reversible work: $W = -nRT \ln\left(\frac{V_2}{V_1}\right)$.
Step $3$: Substitute the values: $W = -2 \times 8.314 \times 300 \times \ln\left(\frac{1}{10^{-3}}\right)$.
Step $4$: Calculate the natural logarithm: $\ln(1000) \approx 6.9078$.
Step $5$: Calculate the final value: $W = -4988.4 \times 6.9078 \approx -34446 \text{ J} = -34.45 \text{ kJ}$.
The closest option is $-34.46 \text{ kJ}$.
$n = 2 \text{ mol}$, $T = 27^{\circ}C = 300 \text{ K}$, $V_1 = 1 \text{ dm}^3 = 10^{-3} \text{ m}^3$, $V_2 = 1 \text{ m}^3$, $R = 8.314 \text{ J K}^{-1} \text{mol}^{-1}$.
Step $2$: Use the formula for isothermal reversible work: $W = -nRT \ln\left(\frac{V_2}{V_1}\right)$.
Step $3$: Substitute the values: $W = -2 \times 8.314 \times 300 \times \ln\left(\frac{1}{10^{-3}}\right)$.
Step $4$: Calculate the natural logarithm: $\ln(1000) \approx 6.9078$.
Step $5$: Calculate the final value: $W = -4988.4 \times 6.9078 \approx -34446 \text{ J} = -34.45 \text{ kJ}$.
The closest option is $-34.46 \text{ kJ}$.
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153
ChemistryMediumMCQMHT CET · 2026
Which of the following processes exhibits a decrease in the internal energy of a system?
A
Isothermal expansion of an ideal gas.
B
Isothermal compression of an ideal gas.
C
Adiabatic expansion of an ideal gas.
D
Adiabatic compression of an ideal gas.
Solution
(C) For an ideal gas, the internal energy $U$ is a function of temperature $T$ only, i.e., $U = f(T)$.
$1$. In isothermal processes, $\Delta T = 0$, so $\Delta U = 0$.
$2$. In adiabatic processes, the first law of thermodynamics is $\Delta U = q + w$. Since $q = 0$, $\Delta U = w$.
$3$. For adiabatic expansion, the gas does work on the surroundings, so $w < 0$. Thus, $\Delta U < 0$, which means internal energy decreases.
$4$. For adiabatic compression, work is done on the gas, so $w > 0$. Thus, $\Delta U > 0$, which means internal energy increases.
Therefore, adiabatic expansion leads to a decrease in internal energy.
$1$. In isothermal processes, $\Delta T = 0$, so $\Delta U = 0$.
$2$. In adiabatic processes, the first law of thermodynamics is $\Delta U = q + w$. Since $q = 0$, $\Delta U = w$.
$3$. For adiabatic expansion, the gas does work on the surroundings, so $w < 0$. Thus, $\Delta U < 0$, which means internal energy decreases.
$4$. For adiabatic compression, work is done on the gas, so $w > 0$. Thus, $\Delta U > 0$, which means internal energy increases.
Therefore, adiabatic expansion leads to a decrease in internal energy.
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154
ChemistryDifficultMCQMHT CET · 2026
Calculate the change in internal energy $(\Delta U)$ if an ideal gas expands from $0.5 \text{ dm}^3$ to $3.5 \text{ dm}^3$ at a constant pressure of $2 \text{ bar}$ by absorbing $1800 \text{ J}$ of heat energy. (in $\text{ J}$)
A
$1200$
B
$600$
C
$1800$
D
$2400$
Solution
(A) Step $1$: Identify the given values.
Heat absorbed $(q)$ = $+1800 \text{ J}$ (positive as heat is absorbed).
Pressure $(P)$ = $2 \text{ bar} = 2 \times 10^5 \text{ Pa}$.
Initial volume $(V_1)$ = $0.5 \text{ dm}^3 = 0.5 \times 10^{-3} \text{ m}^3$.
Final volume $(V_2)$ = $3.5 \text{ dm}^3 = 3.5 \times 10^{-3} \text{ m}^3$.
Step $2$: Calculate the work done $(w)$.
$w = -P \Delta V = -P(V_2 - V_1)$.
$w = -2 \times 10^5 \text{ Pa} \times (3.5 - 0.5) \times 10^{-3} \text{ m}^3$.
$w = -2 \times 10^5 \times 3 \times 10^{-3} \text{ J} = -600 \text{ J}$.
Step $3$: Calculate the change in internal energy $(\Delta U)$ using the first law of thermodynamics.
$\Delta U = q + w$.
$\Delta U = 1800 \text{ J} + (-600 \text{ J}) = 1200 \text{ J}$.
Heat absorbed $(q)$ = $+1800 \text{ J}$ (positive as heat is absorbed).
Pressure $(P)$ = $2 \text{ bar} = 2 \times 10^5 \text{ Pa}$.
Initial volume $(V_1)$ = $0.5 \text{ dm}^3 = 0.5 \times 10^{-3} \text{ m}^3$.
Final volume $(V_2)$ = $3.5 \text{ dm}^3 = 3.5 \times 10^{-3} \text{ m}^3$.
Step $2$: Calculate the work done $(w)$.
$w = -P \Delta V = -P(V_2 - V_1)$.
$w = -2 \times 10^5 \text{ Pa} \times (3.5 - 0.5) \times 10^{-3} \text{ m}^3$.
$w = -2 \times 10^5 \times 3 \times 10^{-3} \text{ J} = -600 \text{ J}$.
Step $3$: Calculate the change in internal energy $(\Delta U)$ using the first law of thermodynamics.
$\Delta U = q + w$.
$\Delta U = 1800 \text{ J} + (-600 \text{ J}) = 1200 \text{ J}$.
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155
ChemistryDifficultMCQMHT CET · 2026
Calculate the $pH$ of a buffer solution containing $0.1 \text{ M } CH_3COOH$ and $0.1 \text{ M } CH_3COONa$ $[K_a = 1.8 \times 10^{-5}]$.
A
$4.745$
B
$7.41$
C
$8.76$
D
$1.00$
Solution
(A) Using the Henderson-Hasselbalch equation: $pH = pK_a + \log \frac{[Salt]}{[Acid]}$.
First, calculate $pK_a$: $pK_a = -\log(K_a) = -\log(1.8 \times 10^{-5}) = 5 - \log(1.8) = 5 - 0.255 = 4.745$.
Since $[Salt] = [CH_3COONa] = 0.1 \text{ M}$ and $[Acid] = [CH_3COOH] = 0.1 \text{ M}$, the ratio $\frac{[Salt]}{[Acid]} = 1$.
Therefore, $pH = 4.745 + \log(1) = 4.745 + 0 = 4.745$.
First, calculate $pK_a$: $pK_a = -\log(K_a) = -\log(1.8 \times 10^{-5}) = 5 - \log(1.8) = 5 - 0.255 = 4.745$.
Since $[Salt] = [CH_3COONa] = 0.1 \text{ M}$ and $[Acid] = [CH_3COOH] = 0.1 \text{ M}$, the ratio $\frac{[Salt]}{[Acid]} = 1$.
Therefore, $pH = 4.745 + \log(1) = 4.745 + 0 = 4.745$.
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156
ChemistryMediumMCQMHT CET · 2026
What happens when $CuSO_4$ is dissolved in water? Identify the correct statement from the following:
A
The anion of the salt reacts with water.
B
Its solution will turn red litmus into blue.
C
The $pH$ of the solution will be less than $7$.
D
The $CuSO_4$ will not undergo hydrolysis reaction.
Solution
(C) $1$. $CuSO_4$ is a salt of a strong acid $(H_2SO_4)$ and a weak base $(Cu(OH)_2)$.
$2$. In water, it dissociates into $Cu^{2+}$ and $SO_4^{2-}$ ions.
$3$. The $Cu^{2+}$ ion undergoes cationic hydrolysis: $Cu^{2+} + 2H_2O \rightleftharpoons Cu(OH)_2 + 2H^+$.
$4$. Due to the production of $H^+$ ions, the solution becomes acidic, resulting in a $pH < 7$.
$2$. In water, it dissociates into $Cu^{2+}$ and $SO_4^{2-}$ ions.
$3$. The $Cu^{2+}$ ion undergoes cationic hydrolysis: $Cu^{2+} + 2H_2O \rightleftharpoons Cu(OH)_2 + 2H^+$.
$4$. Due to the production of $H^+$ ions, the solution becomes acidic, resulting in a $pH < 7$.
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157
ChemistryMediumMCQMHT CET · 2026
What happens when solid $Na_2CO_3$ is dissolved in water?
A
an increase in $[OH^-]$
B
no change in $pH$
C
an increase in $[H^+]$ ions
D
turns blue litmus red
Solution
(A) $Na_2CO_3$ is a salt of a strong base $(NaOH)$ and a weak acid $(H_2CO_3)$.
When dissolved in water, it undergoes anionic hydrolysis: $CO_3^{2-} + H_2O \rightleftharpoons HCO_3^- + OH^-$.
This reaction produces $OH^-$ ions, which increases the concentration of $[OH^-]$ and makes the solution basic $(pH > 7)$.
When dissolved in water, it undergoes anionic hydrolysis: $CO_3^{2-} + H_2O \rightleftharpoons HCO_3^- + OH^-$.
This reaction produces $OH^-$ ions, which increases the concentration of $[OH^-]$ and makes the solution basic $(pH > 7)$.
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158
ChemistryMediumMCQMHT CET · 2026
Which of the following aqueous solutions of salts has $pH < 7$ at $298 \text{ K}$?
A
$KNO_3$
B
$Na_2CO_3$
C
$CH_3COONa$
D
$CuCl_2$
Solution
(D) $1$. $KNO_3$ is a salt of a strong acid $(HNO_3)$ and a strong base $(KOH)$, so its aqueous solution is neutral $(pH = 7)$.
$2$. $Na_2CO_3$ is a salt of a weak acid $(H_2CO_3)$ and a strong base $(NaOH)$, so its aqueous solution is basic $(pH > 7)$.
$3$. $CH_3COONa$ is a salt of a weak acid $(CH_3COOH)$ and a strong base $(NaOH)$, so its aqueous solution is basic $(pH > 7)$.
$4$. $CuCl_2$ is a salt of a strong acid $(HCl)$ and a weak base $(Cu(OH)_2)$. Due to the hydrolysis of the $Cu^{2+}$ ion, the solution becomes acidic $(pH < 7)$.
$2$. $Na_2CO_3$ is a salt of a weak acid $(H_2CO_3)$ and a strong base $(NaOH)$, so its aqueous solution is basic $(pH > 7)$.
$3$. $CH_3COONa$ is a salt of a weak acid $(CH_3COOH)$ and a strong base $(NaOH)$, so its aqueous solution is basic $(pH > 7)$.
$4$. $CuCl_2$ is a salt of a strong acid $(HCl)$ and a weak base $(Cu(OH)_2)$. Due to the hydrolysis of the $Cu^{2+}$ ion, the solution becomes acidic $(pH < 7)$.
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159
ChemistryMediumMCQMHT CET · 2026
Identify from the following salts an example of a salt of a strong acid and a weak base.
A
Ammonium acetate
B
Sodium acetate
C
Copper $(II)$ chloride
D
Sodium carbonate
Solution
(C) Step $1$: Identify the acid and base components for each salt.
Step $2$: $Ammonium \ acetate$ is formed from $CH_3COOH$ (weak acid) and $NH_4OH$ (weak base).
Step $3$: $Sodium \ acetate$ is formed from $CH_3COOH$ (weak acid) and $NaOH$ (strong base).
Step $4$: $Copper(II) \ chloride$ $(CuCl_2)$ is formed from $HCl$ (strong acid) and $Cu(OH)_2$ (weak base).
Step $5$: $Sodium \ carbonate$ is formed from $H_2CO_3$ (weak acid) and $NaOH$ (strong base).
Step $6$: Therefore, $CuCl_2$ is the correct example of a salt of a strong acid and a weak base.
Step $2$: $Ammonium \ acetate$ is formed from $CH_3COOH$ (weak acid) and $NH_4OH$ (weak base).
Step $3$: $Sodium \ acetate$ is formed from $CH_3COOH$ (weak acid) and $NaOH$ (strong base).
Step $4$: $Copper(II) \ chloride$ $(CuCl_2)$ is formed from $HCl$ (strong acid) and $Cu(OH)_2$ (weak base).
Step $5$: $Sodium \ carbonate$ is formed from $H_2CO_3$ (weak acid) and $NaOH$ (strong base).
Step $6$: Therefore, $CuCl_2$ is the correct example of a salt of a strong acid and a weak base.
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160
ChemistryMediumMCQMHT CET · 2026
Which of the following salts undergoes anionic hydrolysis?
A
$CuSO_4$
B
$Na_2CO_3$
C
$NH_4Cl$
D
$FeCl_3$
Solution
(B) $1$. Anionic hydrolysis occurs when the anion of a salt reacts with water to form a basic solution.
$2$. This happens in salts formed from a weak acid and a strong base.
$3$. $Na_2CO_3$ is a salt of a weak acid $(H_2CO_3)$ and a strong base $(NaOH)$.
$4$. The anion $CO_3^{2-}$ reacts with water: $CO_3^{2-} + H_2O \rightleftharpoons HCO_3^- + OH^-$, resulting in anionic hydrolysis.
$5$. $CuSO_4$, $NH_4Cl$, and $FeCl_3$ are salts of strong acids and weak bases, which undergo cationic hydrolysis.
$2$. This happens in salts formed from a weak acid and a strong base.
$3$. $Na_2CO_3$ is a salt of a weak acid $(H_2CO_3)$ and a strong base $(NaOH)$.
$4$. The anion $CO_3^{2-}$ reacts with water: $CO_3^{2-} + H_2O \rightleftharpoons HCO_3^- + OH^-$, resulting in anionic hydrolysis.
$5$. $CuSO_4$, $NH_4Cl$, and $FeCl_3$ are salts of strong acids and weak bases, which undergo cationic hydrolysis.
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161
ChemistryMediumMCQMHT CET · 2026
Which of the following salts does not undergo hydrolysis?
A
Ammonium nitrate
B
Sodium acetate
C
Ammonium cyanide
D
Sodium nitrate
Solution
(D) Step $1$: Hydrolysis occurs when a salt is formed from a weak acid and a weak base, a weak acid and a strong base, or a strong acid and a weak base.
Step $2$: $NaNO_3$ is a salt formed from a strong acid $(HNO_3)$ and a strong base $(NaOH)$.
Step $3$: Salts of strong acids and strong bases do not undergo hydrolysis because their ions do not react with water to change the $pH$ of the solution.
Step $4$: Therefore, $NaNO_3$ does not undergo hydrolysis.
Step $2$: $NaNO_3$ is a salt formed from a strong acid $(HNO_3)$ and a strong base $(NaOH)$.
Step $3$: Salts of strong acids and strong bases do not undergo hydrolysis because their ions do not react with water to change the $pH$ of the solution.
Step $4$: Therefore, $NaNO_3$ does not undergo hydrolysis.
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162
ChemistryMediumMCQMHT CET · 2026
Which of the following aqueous salt solutions has the least $pH$ value?
A
$Na_2CO_3$
B
$CuCl_2$
C
$CH_3COONa$
D
$NaNO_3$
Solution
(B) $(1)$ $Na_2CO_3$ is a salt of a strong base $(NaOH)$ and a weak acid $(H_2CO_3)$, so it is basic $(pH > 7)$.
$(2)$ $CuCl_2$ is a salt of a weak base $(Cu(OH)_2)$ and a strong acid $(HCl)$, so it is acidic $(pH < 7)$.
$(3)$ $CH_3COONa$ is a salt of a strong base $(NaOH)$ and a weak acid $(CH_3COOH)$, so it is basic $(pH > 7)$.
$(4)$ $NaNO_3$ is a salt of a strong base $(NaOH)$ and a strong acid $(HNO_3)$, so it is neutral $(pH = 7)$.
Therefore, $CuCl_2$ has the least $pH$ value.
$(2)$ $CuCl_2$ is a salt of a weak base $(Cu(OH)_2)$ and a strong acid $(HCl)$, so it is acidic $(pH < 7)$.
$(3)$ $CH_3COONa$ is a salt of a strong base $(NaOH)$ and a weak acid $(CH_3COOH)$, so it is basic $(pH > 7)$.
$(4)$ $NaNO_3$ is a salt of a strong base $(NaOH)$ and a strong acid $(HNO_3)$, so it is neutral $(pH = 7)$.
Therefore, $CuCl_2$ has the least $pH$ value.
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163
ChemistryEasyMCQMHT CET · 2026
What is the normal pH of human blood?
A
0
B
9
C
4
D
1
Solution
The normal physiological pH of human blood is approximately 7.4.
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164
ChemistryDifficultMCQMHT CET · 2026
$2.8 \text{ g}$ of $KOH$ is dissolved in a $500 \text{ mL}$ solution at $298 \text{ K}$. What is the $pH$ of the solution? (Molar mass of $KOH = 56 \text{ g/mol}$)
A
$1$
B
$8$
C
$11$
D
$13$
Solution
(D) Step $1$: Calculate the number of moles of $KOH$: $n = \frac{\text{mass}}{\text{molar mass}} = \frac{2.8 \text{ g}}{56 \text{ g/mol}} = 0.05 \text{ mol}$.
Step $2$: Calculate the molarity $(M)$ of the solution: $M = \frac{n}{V(\text{in L})} = \frac{0.05 \text{ mol}}{0.5 \text{ L}} = 0.1 \text{ M}$.
Step $3$: Since $KOH$ is a strong base, $[OH^-] = 0.1 \text{ M} = 10^{-1} \text{ M}$.
Step $4$: Calculate $pOH$: $pOH = -\log[OH^-] = -\log(10^{-1}) = 1$.
Step $5$: Calculate $pH$: $pH = 14 - pOH = 14 - 1 = 13$.
Step $2$: Calculate the molarity $(M)$ of the solution: $M = \frac{n}{V(\text{in L})} = \frac{0.05 \text{ mol}}{0.5 \text{ L}} = 0.1 \text{ M}$.
Step $3$: Since $KOH$ is a strong base, $[OH^-] = 0.1 \text{ M} = 10^{-1} \text{ M}$.
Step $4$: Calculate $pOH$: $pOH = -\log[OH^-] = -\log(10^{-1}) = 1$.
Step $5$: Calculate $pH$: $pH = 14 - pOH = 14 - 1 = 13$.
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165
ChemistryDifficultMCQMHT CET · 2026
$A$ weak monobasic acid is $0.04 \%$ dissociated in $0.25 \text{ M}$ solution. What is the $pH$ of the solution?
A
$5$
B
$4$
C
$3$
D
$10$
Solution
(B) Step $1$: Calculate the concentration of $H^+$ ions.
$[H^+] = C \times \alpha$
Given $C = 0.25 \text{ M}$ and $\alpha = \frac{0.04}{100} = 4 \times 10^{-4}$.
$[H^+] = 0.25 \times 4 \times 10^{-4} = 1 \times 10^{-4} \text{ M}$.
Step $2$: Calculate the $pH$ of the solution.
$pH = -\log[H^+]$
$pH = -\log(10^{-4}) = 4$.
$[H^+] = C \times \alpha$
Given $C = 0.25 \text{ M}$ and $\alpha = \frac{0.04}{100} = 4 \times 10^{-4}$.
$[H^+] = 0.25 \times 4 \times 10^{-4} = 1 \times 10^{-4} \text{ M}$.
Step $2$: Calculate the $pH$ of the solution.
$pH = -\log[H^+]$
$pH = -\log(10^{-4}) = 4$.
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166
ChemistryDifficultMCQMHT CET · 2026
The dissociation constant of a weak acid $HA$ is $1.5 \times 10^{-5}$. Find the percent dissociation of a solution containing $0.2 \text{ moles}$ of $HA$ in $2 \text{ liters}$ of solution. (in $\%$)
A
$1.22$
B
$0.86$
C
$1.50$
D
$2.10$
Solution
(A) Step $1$: Calculate the molar concentration $C$ of the solution.
$C = \frac{n}{V} = \frac{0.2 \text{ mol}}{2 \text{ L}} = 0.1 \text{ M}$.
Step $2$: Use the formula for the degree of dissociation $\alpha$ for a weak acid.
$\alpha = \sqrt{\frac{K_a}{C}} = \sqrt{\frac{1.5 \times 10^{-5}}{0.1}} = \sqrt{1.5 \times 10^{-4}} = 1.2247 \times 10^{-2}$.
Step $3$: Convert the degree of dissociation to percent dissociation.
$\text{Percent dissociation} = \alpha \times 100 = 1.2247 \times 10^{-2} \times 100 = 1.2247 \% \approx 1.22 \%$.
$C = \frac{n}{V} = \frac{0.2 \text{ mol}}{2 \text{ L}} = 0.1 \text{ M}$.
Step $2$: Use the formula for the degree of dissociation $\alpha$ for a weak acid.
$\alpha = \sqrt{\frac{K_a}{C}} = \sqrt{\frac{1.5 \times 10^{-5}}{0.1}} = \sqrt{1.5 \times 10^{-4}} = 1.2247 \times 10^{-2}$.
Step $3$: Convert the degree of dissociation to percent dissociation.
$\text{Percent dissociation} = \alpha \times 100 = 1.2247 \times 10^{-2} \times 100 = 1.2247 \% \approx 1.22 \%$.
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167
ChemistryDifficultMCQMHT CET · 2026
Calculate the $pOH$ of a $0.001 \text{ M } HCl$ solution.
A
$11$
B
$10$
C
$12$
D
$3$
Solution
(A) Step $1$: $HCl$ is a strong acid and dissociates completely as $HCl \rightarrow H^+ + Cl^-$.
Step $2$: The concentration of $[H^+]$ is $0.001 \text{ M} = 10^{-3} \text{ M}$.
Step $3$: Calculate $pH$ using the formula $pH = -\log[H^+] = -\log(10^{-3}) = 3$.
Step $4$: Use the relation $pH + pOH = 14$ at $25^\circ \text{C}$.
Step $5$: $pOH = 14 - 3 = 11$.
Step $2$: The concentration of $[H^+]$ is $0.001 \text{ M} = 10^{-3} \text{ M}$.
Step $3$: Calculate $pH$ using the formula $pH = -\log[H^+] = -\log(10^{-3}) = 3$.
Step $4$: Use the relation $pH + pOH = 14$ at $25^\circ \text{C}$.
Step $5$: $pOH = 14 - 3 = 11$.
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168
ChemistryDifficultMCQMHT CET · 2026
What is the $pH$ of the resulting solution when $20 \text{ mL}$ of $\frac{M}{10} \text{ NaOH}$ and $10 \text{ mL}$ of $\frac{M}{10} \text{ H}_2\text{SO}_4$ are mixed together?
A
$0$
B
$2$
C
$7$
D
$10$
Solution
(C) Step $1$: Calculate moles of $OH^-$ ions from $\text{NaOH}$.
$\text{Moles of } OH^- = \text{Molarity} \times \text{Volume (in mL)} = 0.1 \text{ M} \times 20 \text{ mL} = 2 \text{ mmol}$.
Step $2$: Calculate moles of $H^+$ ions from $\text{H}_2\text{SO}_4$.
$\text{Moles of } H^+ = 2 \times \text{Molarity} \times \text{Volume (in mL)} = 2 \times 0.1 \text{ M} \times 10 \text{ mL} = 2 \text{ mmol}$.
Step $3$: Compare the moles of $H^+$ and $OH^-$.
Since the moles of $H^+$ $(2 \text{ mmol})$ are equal to the moles of $OH^-$ $(2 \text{ mmol})$, the solution is neutral.
Step $4$: Determine the $pH$.
For a neutral solution at $25^{\circ}\text{C}$, the $pH = 7$.
$\text{Moles of } OH^- = \text{Molarity} \times \text{Volume (in mL)} = 0.1 \text{ M} \times 20 \text{ mL} = 2 \text{ mmol}$.
Step $2$: Calculate moles of $H^+$ ions from $\text{H}_2\text{SO}_4$.
$\text{Moles of } H^+ = 2 \times \text{Molarity} \times \text{Volume (in mL)} = 2 \times 0.1 \text{ M} \times 10 \text{ mL} = 2 \text{ mmol}$.
Step $3$: Compare the moles of $H^+$ and $OH^-$.
Since the moles of $H^+$ $(2 \text{ mmol})$ are equal to the moles of $OH^-$ $(2 \text{ mmol})$, the solution is neutral.
Step $4$: Determine the $pH$.
For a neutral solution at $25^{\circ}\text{C}$, the $pH = 7$.
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169
ChemistryDifficultMCQMHT CET · 2026
What is the molarity of $H_2SO_4$ solution having $pH = 4$?
A
$4 \times 10^{-1} \text{ M}$
B
$4 \times 10^{-2} \text{ M}$
C
$5 \times 10^{-5} \text{ M}$
D
$1 \times 10^{-4} \text{ M}$
Solution
(C) The concentration of hydrogen ions is given by $[H^+] = 10^{-pH} = 10^{-4} \text{ M}$.
Since $H_2SO_4$ is a strong diprotic acid, it dissociates as $H_2SO_4 \rightarrow 2H^+ + SO_4^{2-}$.
Therefore, the molarity of the $H_2SO_4$ solution is $[H_2SO_4] = \frac{[H^+]}{2}$.
$[H_2SO_4] = \frac{10^{-4}}{2} \text{ M} = 0.5 \times 10^{-4} \text{ M} = 5 \times 10^{-5} \text{ M}$.
Since $H_2SO_4$ is a strong diprotic acid, it dissociates as $H_2SO_4 \rightarrow 2H^+ + SO_4^{2-}$.
Therefore, the molarity of the $H_2SO_4$ solution is $[H_2SO_4] = \frac{[H^+]}{2}$.
$[H_2SO_4] = \frac{10^{-4}}{2} \text{ M} = 0.5 \times 10^{-4} \text{ M} = 5 \times 10^{-5} \text{ M}$.
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170
ChemistryDifficultMCQMHT CET · 2026
An organic monobasic acid has a dissociation constant of $1.96 \times 10^{-8}$. What is its percentage dissociation in a $0.01 \text{ M}$ solution (in $\%$)?
A
$40$
B
$0.14$
C
$4$
D
$0.19$
Solution
(B) For a weak monobasic acid, the degree of dissociation $\alpha$ is given by $\alpha = \sqrt{K_a / C}$.
Given $K_a = 1.96 \times 10^{-8}$ and $C = 0.01 \text{ M}$.
$\alpha = \sqrt{\frac{1.96 \times 10^{-8}}{0.01}} = \sqrt{1.96 \times 10^{-6}} = 1.4 \times 10^{-3}$.
Percentage dissociation = $\alpha \times 100 = 1.4 \times 10^{-3} \times 100 = 0.14 \%$.
Given $K_a = 1.96 \times 10^{-8}$ and $C = 0.01 \text{ M}$.
$\alpha = \sqrt{\frac{1.96 \times 10^{-8}}{0.01}} = \sqrt{1.96 \times 10^{-6}} = 1.4 \times 10^{-3}$.
Percentage dissociation = $\alpha \times 100 = 1.4 \times 10^{-3} \times 100 = 0.14 \%$.
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171
ChemistryDifficultMCQMHT CET · 2026
What is the $pH$ of a solution containing $5 \times 10^{-4} \text{ M } H^+$ ions?
A
$3.301$
B
$4.301$
C
$2.301$
D
$3.699$
Solution
(A) The formula for $pH$ is $pH = -\log[H^+]$.
Given $[H^+] = 5 \times 10^{-4} \text{ M}$.
$pH = -\log(5 \times 10^{-4})$.
$pH = -(\log 5 + \log 10^{-4})$.
$pH = -(\log 5 - 4)$.
$pH = 4 - \log 5$.
Since $\log 5 \approx 0.699$,
$pH = 4 - 0.699 = 3.301$.
Given $[H^+] = 5 \times 10^{-4} \text{ M}$.
$pH = -\log(5 \times 10^{-4})$.
$pH = -(\log 5 + \log 10^{-4})$.
$pH = -(\log 5 - 4)$.
$pH = 4 - \log 5$.
Since $\log 5 \approx 0.699$,
$pH = 4 - 0.699 = 3.301$.
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172
ChemistryEasyMCQMHT CET · 2026
What is the normal $pH$ range of human blood?
A
$7.0$ to $7.2$
B
$7.35$ to $7.45$
C
$7.5$ to $7.8$
D
$6.8$ to $7.0$
Solution
(B) The normal $pH$ range of human blood is tightly regulated between $7.35$ and $7.45$ to maintain physiological homeostasis. Any value outside this range can lead to conditions like acidosis or alkalosis.
Therefore, the correct option is $B$.
Therefore, the correct option is $B$.
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173
ChemistryEasyMCQMHT CET · 2026
Which of the following ions are responsible for the hardness of water?
A
$Ag^+, Au^{3+}$ ions
B
$Hg^{2+}, Pb^{2+}$ ions
C
$Ca^{2+}$ and $Mg^{2+}$ ions
D
$As^{3+}, CN^-$ ions
Solution
(C) Hardness of water is primarily caused by the presence of dissolved calcium $(Ca^{2+})$ and magnesium $(Mg^{2+})$ ions in the water. These ions react with soap to form insoluble precipitates, preventing the formation of lather.
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174
ChemistryMediumMCQMHT CET · 2026
What is the role of glycerol in the decomposition of $H_2O_2$?
A
It increases the reaction rate.
B
It reduces the reaction rate.
C
It increases the enthalpy of the reaction.
D
It increases the rate of collisions between reactant molecules.
Solution
(B) Step $1$: The decomposition of hydrogen peroxide $(H_2O_2)$ is a spontaneous reaction that is often catalyzed by impurities or light.
Step $2$: Glycerol acts as a negative catalyst or stabilizer in this reaction.
Step $3$: By adding glycerol, the rate of decomposition of $H_2O_2$ is significantly decreased, thereby stabilizing the solution.
Step $4$: Therefore, glycerol reduces the reaction rate.
Step $2$: Glycerol acts as a negative catalyst or stabilizer in this reaction.
Step $3$: By adding glycerol, the rate of decomposition of $H_2O_2$ is significantly decreased, thereby stabilizing the solution.
Step $4$: Therefore, glycerol reduces the reaction rate.
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175
ChemistryEasyMCQMHT CET · 2026
Which of the following is obtained when dihydrogen reacts with the alkali metals at high temperature?
A
Alkoxides
B
Metal hydrides
C
Peroxides
D
Metal nitrites
Solution
(B) Step $1$: Alkali metals $(M)$ react with dihydrogen $(H_2)$ at high temperatures to form ionic or saline hydrides.
Step $2$: The general chemical equation for this reaction is: $2M(s) + H_2(g) \xrightarrow{\Delta} 2MH(s)$, where $M$ represents an alkali metal.
Step $3$: Therefore, the product obtained is a metal hydride.
Step $2$: The general chemical equation for this reaction is: $2M(s) + H_2(g) \xrightarrow{\Delta} 2MH(s)$, where $M$ represents an alkali metal.
Step $3$: Therefore, the product obtained is a metal hydride.
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176
ChemistryEasyMCQMHT CET · 2026
Which of the following is the most abundant element in the universe?
A
$H$
B
$O$
C
$Ne$
D
$C$
Solution
(A) $1$. The universe is composed primarily of light elements formed during the Big Bang.
$2$. Hydrogen $(H)$ is the simplest and most abundant element, accounting for approximately $75\%$ of the baryonic mass of the universe.
$3$. Helium $(He)$ is the second most abundant, while other elements like Oxygen $(O)$, Carbon $(C)$, and Neon $(Ne)$ exist in much smaller quantities.
$2$. Hydrogen $(H)$ is the simplest and most abundant element, accounting for approximately $75\%$ of the baryonic mass of the universe.
$3$. Helium $(He)$ is the second most abundant, while other elements like Oxygen $(O)$, Carbon $(C)$, and Neon $(Ne)$ exist in much smaller quantities.
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177
ChemistryEasyMCQMHT CET · 2026
The most abundant element in the universe is
A
$Hydrogen$
B
$Sodium$
C
$Magnesium$
D
$Potassium$
Solution
(A) $1$. The universe is primarily composed of light elements formed during the Big Bang.
$2$. $Hydrogen$ $(H)$ accounts for approximately $75\%$ of the elemental mass of the universe.
$3$. $Helium$ $(He)$ is the second most abundant, while all other elements are present in much smaller quantities.
$4$. Therefore, $Hydrogen$ is the most abundant element.
$2$. $Hydrogen$ $(H)$ accounts for approximately $75\%$ of the elemental mass of the universe.
$3$. $Helium$ $(He)$ is the second most abundant, while all other elements are present in much smaller quantities.
$4$. Therefore, $Hydrogen$ is the most abundant element.
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178
ChemistryEasyMCQMHT CET · 2026
Which of the following statements is $NOT$ correct regarding dihydrogen $(H_2)$?
A
It is a colorless and tasteless gas.
B
It burns with a pale blue flame.
C
It is a polar water-soluble gas.
D
It is lighter than air.
Solution
(C) $1$. Dihydrogen $(H_2)$ is a non-polar molecule because the electronegativity difference between the two hydrogen atoms is zero.
$2$. Due to its non-polar nature, it has very low solubility in water.
$3$. It is a colorless, odorless, and tasteless gas.
$4$. It is lighter than air and burns with a pale blue flame.
$5$. Therefore, the statement that it is a polar water-soluble gas is incorrect.
$2$. Due to its non-polar nature, it has very low solubility in water.
$3$. It is a colorless, odorless, and tasteless gas.
$4$. It is lighter than air and burns with a pale blue flame.
$5$. Therefore, the statement that it is a polar water-soluble gas is incorrect.
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179
ChemistryEasyMCQMHT CET · 2026
Which of the following statements is false regarding hydrogen?
A
It has two isotopes.
B
In periodic table hydrogen is placed separately above group $1$.
C
When one electron is lost it forms a proton.
D
It does not exist freely.
Solution
(A) Step $1$: Hydrogen has three naturally occurring isotopes: protium $(^1H)$, deuterium ($^2H$ or $D$), and tritium ($^3H$ or $T$).
Step $2$: Statement $(A)$ claims it has two isotopes, which is false.
Step $3$: Statement $(B)$ is true as hydrogen's position is unique due to its properties.
Step $4$: Statement $(C)$ is true because $H - e^- \rightarrow H^+$, which is a proton.
Step $5$: Statement $(D)$ is true as hydrogen exists as a diatomic molecule $(H_2)$ in nature, not as a free atom.
Step $2$: Statement $(A)$ claims it has two isotopes, which is false.
Step $3$: Statement $(B)$ is true as hydrogen's position is unique due to its properties.
Step $4$: Statement $(C)$ is true because $H - e^- \rightarrow H^+$, which is a proton.
Step $5$: Statement $(D)$ is true as hydrogen exists as a diatomic molecule $(H_2)$ in nature, not as a free atom.
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180
ChemistryMediumMCQMHT CET · 2026
Which alkaline earth metal carbonate decomposes most easily?
A
$MgCO_3$
B
$CaCO_3$
C
$SrCO_3$
D
$BaCO_3$
Solution
(A) The thermal stability of alkaline earth metal carbonates increases down the group as the electropositive character of the metal increases.
$Mg^{2+}$ has the smallest ionic radius among the given options, which leads to high polarising power.
This high polarising power distorts the electron cloud of the carbonate ion $(CO_3^{2-})$ more effectively, weakening the $C-O$ bond.
Therefore, $MgCO_3$ requires the least amount of energy to decompose compared to $CaCO_3$, $SrCO_3$, and $BaCO_3$.
$Mg^{2+}$ has the smallest ionic radius among the given options, which leads to high polarising power.
This high polarising power distorts the electron cloud of the carbonate ion $(CO_3^{2-})$ more effectively, weakening the $C-O$ bond.
Therefore, $MgCO_3$ requires the least amount of energy to decompose compared to $CaCO_3$, $SrCO_3$, and $BaCO_3$.
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181
ChemistryEasyMCQMHT CET · 2026
Which of the following compounds is formed when chlorine gas reacts with dry slaked lime $Ca(OH)_2$?
A
$CaOCl_2$
B
$CaO_2Cl_2$
C
$CaOCl$
D
$CaCl_2$
Solution
(A) The reaction between chlorine gas and dry slaked lime $(Ca(OH)_2)$ produces bleaching powder $(CaOCl_2)$ and water $(H_2O)$.
The chemical equation is:
$Ca(OH)_2 + Cl_2 \rightarrow CaOCl_2 + H_2O$
Therefore, the compound formed is $CaOCl_2$.
The chemical equation is:
$Ca(OH)_2 + Cl_2 \rightarrow CaOCl_2 + H_2O$
Therefore, the compound formed is $CaOCl_2$.
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182
ChemistryMediumMCQMHT CET · 2026
Choose the false statement regarding the properties of sodium hydroxide $(NaOH)$.
A
It is highly water soluble.
B
It absorbs atmospheric $CO_2$ to form $Na_2CO_3$.
C
The solution of $NaOH$ turns blue litmus red.
D
$NaOH$ is a white deliquescent solid.
Solution
(C) Step $1$: Sodium hydroxide $(NaOH)$ is a strong base.
Step $2$: Strong bases turn red litmus blue, not blue litmus red.
Step $3$: Therefore, the statement that $NaOH$ turns blue litmus red is false.
Step $4$: $NaOH$ is highly soluble in water, absorbs $CO_2$ from the atmosphere to form $Na_2CO_3$, and is a white deliquescent solid, making those statements true.
Step $2$: Strong bases turn red litmus blue, not blue litmus red.
Step $3$: Therefore, the statement that $NaOH$ turns blue litmus red is false.
Step $4$: $NaOH$ is highly soluble in water, absorbs $CO_2$ from the atmosphere to form $Na_2CO_3$, and is a white deliquescent solid, making those statements true.
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183
ChemistryMediumMCQMHT CET · 2026
Identify the elements of alkali metals and alkaline earth metals respectively having the highest $1^{st}$ ionization enthalpy.
A
$K$ and $Ca$
B
$Li$ and $Be$
C
$Rb$ and $Ba$
D
$K$ and $Sr$
Solution
(B) $1$. Ionization enthalpy decreases down a group as the atomic size increases and the valence electrons are further from the nucleus.
$2$. Alkali metals $(Group \ 1)$ follow the order: $Li > Na > K > Rb > Cs$. Thus, $Li$ has the highest $1^{st}$ ionization enthalpy.
$3$. Alkaline earth metals $(Group \ 2)$ follow the order: $Be > Mg > Ca > Sr > Ba$. Thus, $Be$ has the highest $1^{st}$ ionization enthalpy.
$4$. Therefore, the correct pair is $Li$ and $Be$.
$2$. Alkali metals $(Group \ 1)$ follow the order: $Li > Na > K > Rb > Cs$. Thus, $Li$ has the highest $1^{st}$ ionization enthalpy.
$3$. Alkaline earth metals $(Group \ 2)$ follow the order: $Be > Mg > Ca > Sr > Ba$. Thus, $Be$ has the highest $1^{st}$ ionization enthalpy.
$4$. Therefore, the correct pair is $Li$ and $Be$.
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184
ChemistryEasyMCQMHT CET · 2026
Identify the alkali metal having the smallest atomic size from the following.
A
Sodium
B
Potassium
C
Rubidium
D
Caesium
Solution
(A) $1$. Alkali metals belong to Group $1$ of the periodic table.
$2$. The order of elements in Group $1$ from top to bottom is $Li, Na, K, Rb, Cs, Fr$.
$3$. Atomic size increases down a group due to the addition of new electron shells.
$4$. Among the given options, $Na$ $(Z=11)$ is the highest in the group, followed by $K$ $(Z=19)$, $Rb$ $(Z=37)$, and $Cs$ $(Z=55)$.
$5$. Therefore, $Na$ has the smallest atomic size among the given options.
$2$. The order of elements in Group $1$ from top to bottom is $Li, Na, K, Rb, Cs, Fr$.
$3$. Atomic size increases down a group due to the addition of new electron shells.
$4$. Among the given options, $Na$ $(Z=11)$ is the highest in the group, followed by $K$ $(Z=19)$, $Rb$ $(Z=37)$, and $Cs$ $(Z=55)$.
$5$. Therefore, $Na$ has the smallest atomic size among the given options.
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185
ChemistryMediumMCQMHT CET · 2026
Which of the following metal chlorides does not contain water of crystallisation?
A
$LiCl$
B
$MgCl_2$
C
$CsCl$
D
$BeCl_2$
Solution
(C) $1$. Water of crystallisation is the fixed number of water molecules present in one formula unit of a salt.
$2$. Small, highly charged cations like $Li^+$, $Mg^{2+}$, and $Be^{2+}$ have high hydration energy and tend to form hydrated salts (e.g., $LiCl \cdot H_2O$, $MgCl_2 \cdot 6H_2O$, $BeCl_2 \cdot 4H_2O$).
$3$. $Cs^+$ is a large cation with low charge density, resulting in low hydration energy. Therefore, $CsCl$ crystallises as an anhydrous salt.
$4$. Thus, $CsCl$ does not contain water of crystallisation.
$2$. Small, highly charged cations like $Li^+$, $Mg^{2+}$, and $Be^{2+}$ have high hydration energy and tend to form hydrated salts (e.g., $LiCl \cdot H_2O$, $MgCl_2 \cdot 6H_2O$, $BeCl_2 \cdot 4H_2O$).
$3$. $Cs^+$ is a large cation with low charge density, resulting in low hydration energy. Therefore, $CsCl$ crystallises as an anhydrous salt.
$4$. Thus, $CsCl$ does not contain water of crystallisation.
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186
ChemistryEasyMCQMHT CET · 2026
What is the general formula of halide salts of alkali metals if $M$ is a metal atom and $X$ is a halogen atom?
A
$MX$
B
$MX_2$
C
$M_3X_2$
D
$M_2X_3$
Solution
(A) $1$. Alkali metals belong to Group $1$ of the periodic table and have a valence shell electronic configuration of $ns^1$.
$2$. Alkali metals lose one electron to form a unipositive ion, $M^+$.
$3$. Halogens belong to Group $17$ and have a valence shell electronic configuration of $ns^2 np^5$, requiring one electron to complete their octet, forming a halide ion, $X^-$.
$4$. Combining $M^+$ and $X^-$ in a $1:1$ ratio to maintain electrical neutrality results in the formula $MX$.
$2$. Alkali metals lose one electron to form a unipositive ion, $M^+$.
$3$. Halogens belong to Group $17$ and have a valence shell electronic configuration of $ns^2 np^5$, requiring one electron to complete their octet, forming a halide ion, $X^-$.
$4$. Combining $M^+$ and $X^-$ in a $1:1$ ratio to maintain electrical neutrality results in the formula $MX$.
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187
ChemistryEasyMCQMHT CET · 2026
Which of the following elements forms a superoxide when it reacts with oxygen?
A
$Sodium$
B
$Potassium$
C
$Lithium$
D
$Magnesium$
Solution
(B) $1$. Alkali metals react with oxygen to form different types of oxides depending on their size.
$2$. $Lithium$ forms only monoxide $(Li_2O)$.
$3$. $Sodium$ forms peroxide $(Na_2O_2)$.
$4$. $Potassium$, $Rubidium$, and $Cesium$ form superoxides $(MO_2)$ due to the stabilization of the large anion by the large cation.
$5$. Therefore, $Potassium$ is the correct answer.
$2$. $Lithium$ forms only monoxide $(Li_2O)$.
$3$. $Sodium$ forms peroxide $(Na_2O_2)$.
$4$. $Potassium$, $Rubidium$, and $Cesium$ form superoxides $(MO_2)$ due to the stabilization of the large anion by the large cation.
$5$. Therefore, $Potassium$ is the correct answer.
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188
ChemistryEasyMCQMHT CET · 2026
What is the chemical formula of Tin $(IV)$ oxide?
A
$Sn_2O_3$
B
$SnO_2$
C
$SnO_4$
D
$SnO$
Solution
(B) Step $1$: Identify the symbols and valencies of the elements. Tin $(IV)$ has a valency of $+4$ $(Sn^{4+})$ and Oxide has a valency of $-2$ $(O^{2-})$.
Step $2$: Use the criss-cross method to determine the formula. The subscript of $Sn$ becomes $2$ and the subscript of $O$ becomes $4$, resulting in $Sn_2O_4$.
Step $3$: Simplify the ratio of the subscripts to the lowest whole number. Dividing both by $2$ gives $SnO_2$.
Step $2$: Use the criss-cross method to determine the formula. The subscript of $Sn$ becomes $2$ and the subscript of $O$ becomes $4$, resulting in $Sn_2O_4$.
Step $3$: Simplify the ratio of the subscripts to the lowest whole number. Dividing both by $2$ gives $SnO_2$.
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189
ChemistryEasyMCQMHT CET · 2026
Which of the following is $NOT$ a polymorphic form of silica $(SiO_2)$?
A
Cristobalite
B
$ \alpha $-quartz
C
$ \beta $-quartz
D
Aragonite
Solution
(D) $1$. Silica $(SiO_2)$ exists in several polymorphic forms such as quartz, tridymite, and cristobalite.
$2$. Quartz exists in different temperature-dependent forms, specifically $ \alpha $-quartz and $ \beta $-quartz.
$3$. Aragonite is a polymorphic form of calcium carbonate $(CaCO_3)$, not silica.
$4$. Therefore, Aragonite is not a polymorphic form of silica.
$2$. Quartz exists in different temperature-dependent forms, specifically $ \alpha $-quartz and $ \beta $-quartz.
$3$. Aragonite is a polymorphic form of calcium carbonate $(CaCO_3)$, not silica.
$4$. Therefore, Aragonite is not a polymorphic form of silica.
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190
ChemistryEasyMCQMHT CET · 2026
Identify the catalyst used to transform carbon monoxide from water gas into carbon dioxide.
A
$K_2Cr_2O_7$
B
$Na_2CrO_4$
C
$FeCrO_4$
D
$Ni$
Solution
(C) The reaction is known as the water-gas shift reaction: $CO(g) + H_2O(g) \xrightarrow{FeCrO_4} CO_2(g) + H_2(g)$.
In this process, carbon monoxide from water gas is reacted with steam in the presence of an iron chromate $(FeCrO_4)$ catalyst to produce carbon dioxide and hydrogen.
In this process, carbon monoxide from water gas is reacted with steam in the presence of an iron chromate $(FeCrO_4)$ catalyst to produce carbon dioxide and hydrogen.
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191
ChemistryEasyMCQMHT CET · 2026
What is the nature of $Al_2O_3$?
A
Acidic
B
Basic
C
Amphoteric
D
Neutral
Solution
(C) $Al_2O_3$ is an amphoteric oxide because it reacts with both acids and bases to form salt and water.
Reaction with acid: $Al_2O_3 + 6HCl \rightarrow 2AlCl_3 + 3H_2O$
Reaction with base: $Al_2O_3 + 2NaOH + 3H_2O \rightarrow 2Na[Al(OH)_4]$
Reaction with acid: $Al_2O_3 + 6HCl \rightarrow 2AlCl_3 + 3H_2O$
Reaction with base: $Al_2O_3 + 2NaOH + 3H_2O \rightarrow 2Na[Al(OH)_4]$
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192
ChemistryEasyMCQMHT CET · 2026
Which of the following oxides is amphoteric in nature?
A
$BaO$
B
$N_2O_5$
C
$CO_2$
D
$Al_2O_3$
Solution
(D) $1$. Amphoteric oxides are those that can react with both acids and bases to form salt and water.
$2$. $BaO$ is a metallic oxide and is basic in nature.
$3$. $N_2O_5$ and $CO_2$ are non-metallic oxides and are acidic in nature.
$4$. $Al_2O_3$ reacts with both $HCl$ (acid) and $NaOH$ (base), hence it is amphoteric in nature.
$2$. $BaO$ is a metallic oxide and is basic in nature.
$3$. $N_2O_5$ and $CO_2$ are non-metallic oxides and are acidic in nature.
$4$. $Al_2O_3$ reacts with both $HCl$ (acid) and $NaOH$ (base), hence it is amphoteric in nature.
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193
ChemistryMediumMCQMHT CET · 2026
What type of isomerism is exhibited by neopentane $(C_5H_{12})$?
A
position isomerism
B
chain isomerism
C
geometrical isomerism
D
tautomerism
Solution
(B) $1$. Neopentane $(2,2-\text{dimethylpropane})$ has the molecular formula $C_5H_{12}$.
$2$. It is an isomer of $n-\text{pentane}$ and isopentane.
$3$. These isomers differ in the arrangement of the carbon chain (straight chain vs. branched chain).
$4$. Therefore, neopentane exhibits chain isomerism.
$2$. It is an isomer of $n-\text{pentane}$ and isopentane.
$3$. These isomers differ in the arrangement of the carbon chain (straight chain vs. branched chain).
$4$. Therefore, neopentane exhibits chain isomerism.
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194
ChemistryDifficultMCQMHT CET · 2026
How many chiral carbon atoms are present in $2\text{-Iodo-}3,4,5\text{-trimethylhexane}$?
A
$2$
B
$3$
C
$4$
D
$1$
Solution
(B) $1$. The structure of $2\text{-Iodo-}3,4,5\text{-trimethylhexane}$ is $CH_3-CH(I)-CH(CH_3)-CH(CH_3)-CH(CH_3)-CH_3$.
$2$. $A$ chiral carbon atom is a carbon atom bonded to four different groups.
$3$. Let us examine each carbon atom:
- $C2$: Bonded to $-H, -I, -CH_3$, and $-CH(CH_3)CH(CH_3)CH(CH_3)CH_3$. This is chiral.
- $C3$: Bonded to $-H, -CH_3, -CH(I)CH_3$, and $-CH(CH_3)CH(CH_3)CH_3$. This is chiral.
- $C4$: Bonded to $-H, -CH_3, -CH(CH_3)CH(I)CH_3$, and $-CH(CH_3)CH_3$. This is chiral.
- $C5$: Bonded to $-H, -CH_3, -CH_3$, and $-CH(CH_3)CH(CH_3)CH(I)CH_3$. Since it is bonded to two identical $-CH_3$ groups, it is achiral.
$4$. Thus, there are $3$ chiral carbon atoms $(C2, C3, C4)$.
$2$. $A$ chiral carbon atom is a carbon atom bonded to four different groups.
$3$. Let us examine each carbon atom:
- $C2$: Bonded to $-H, -I, -CH_3$, and $-CH(CH_3)CH(CH_3)CH(CH_3)CH_3$. This is chiral.
- $C3$: Bonded to $-H, -CH_3, -CH(I)CH_3$, and $-CH(CH_3)CH(CH_3)CH_3$. This is chiral.
- $C4$: Bonded to $-H, -CH_3, -CH(CH_3)CH(I)CH_3$, and $-CH(CH_3)CH_3$. This is chiral.
- $C5$: Bonded to $-H, -CH_3, -CH_3$, and $-CH(CH_3)CH(CH_3)CH(I)CH_3$. Since it is bonded to two identical $-CH_3$ groups, it is achiral.
$4$. Thus, there are $3$ chiral carbon atoms $(C2, C3, C4)$.
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195
ChemistryDifficultMCQMHT CET · 2026
What is the number of chiral carbon atoms present in $2$-Bromo-$3,4,5$-trimethylhexane?
A
$1$
B
$2$
C
$3$
D
$4$
Solution
(D) The structure of $2$-Bromo-$3,4,5$-trimethylhexane is $CH_3-CH(Br)-CH(CH_3)-CH(CH_3)-CH(CH_3)-CH_3$.
Identify the chiral carbons (carbons attached to four different groups):
$(1)$ $C_2$: Attached to $-H$, $-Br$, $-CH_3$, and $-CH(CH_3)CH(CH_3)CH_2CH_3$. This is chiral.
$(2)$ $C_3$: Attached to $-H$, $-CH_3$, $-CH(Br)CH_3$, and $-CH(CH_3)CH_2CH_3$. This is chiral.
$(3)$ $C_4$: Attached to $-H$, $-CH_3$, $-CH(CH_3)CH(Br)CH_3$, and $-CH_2CH_3$. This is chiral.
$(4)$ $C_5$: Attached to $-H$, $-CH_3$, $-CH_2CH_3$, and $-CH(CH_3)CH(CH_3)CH(Br)CH_3$. This is chiral.
All four carbons at positions $2, 3, 4,$ and $5$ are chiral.
Therefore, the total number of chiral carbon atoms is $4$.
Identify the chiral carbons (carbons attached to four different groups):
$(1)$ $C_2$: Attached to $-H$, $-Br$, $-CH_3$, and $-CH(CH_3)CH(CH_3)CH_2CH_3$. This is chiral.
$(2)$ $C_3$: Attached to $-H$, $-CH_3$, $-CH(Br)CH_3$, and $-CH(CH_3)CH_2CH_3$. This is chiral.
$(3)$ $C_4$: Attached to $-H$, $-CH_3$, $-CH(CH_3)CH(Br)CH_3$, and $-CH_2CH_3$. This is chiral.
$(4)$ $C_5$: Attached to $-H$, $-CH_3$, $-CH_2CH_3$, and $-CH(CH_3)CH(CH_3)CH(Br)CH_3$. This is chiral.
All four carbons at positions $2, 3, 4,$ and $5$ are chiral.
Therefore, the total number of chiral carbon atoms is $4$.
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196
ChemistryMediumMCQMHT CET · 2026
What is the number of chiral carbon atoms in $2$-chlorobutane?
A
$4$
B
$3$
C
$2$
D
$1$
Solution
(D) The structure of $2$-chlorobutane is $CH_3-CHCl-CH_2-CH_3$.
$A$ chiral carbon atom is a carbon atom bonded to four different groups.
In $2$-chlorobutane, the carbon at position $2$ is bonded to:
$1$. $A$ hydrogen atom $(-H)$
$2$. $A$ chlorine atom $(-Cl)$
$3$. $A$ methyl group $(-CH_3)$
$4$. An ethyl group $(-CH_2CH_3)$
Since all four groups are different, the carbon at position $2$ is chiral.
There is only $1$ chiral carbon atom in the molecule.
$A$ chiral carbon atom is a carbon atom bonded to four different groups.
In $2$-chlorobutane, the carbon at position $2$ is bonded to:
$1$. $A$ hydrogen atom $(-H)$
$2$. $A$ chlorine atom $(-Cl)$
$3$. $A$ methyl group $(-CH_3)$
$4$. An ethyl group $(-CH_2CH_3)$
Since all four groups are different, the carbon at position $2$ is chiral.
There is only $1$ chiral carbon atom in the molecule.
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197
ChemistryMediumMCQMHT CET · 2026
Find the number of structural isomers possible for hexane $(C_6H_{14})$?
A
Two
B
Four
C
Three
D
Five
Solution
(D) The structural isomers of hexane $(C_6H_{14})$ are:
$1$. $n$-hexane: $CH_3-CH_2-CH_2-CH_2-CH_2-CH_3$
$2$. $2$-methylpentane: $CH_3-CH(CH_3)-CH_2-CH_2-CH_3$
$3$. $3$-methylpentane: $CH_3-CH_2-CH(CH_3)-CH_2-CH_3$
$4$. $2,2$-dimethylbutane: $CH_3-C(CH_3)_2-CH_2-CH_3$
$5$. $2,3$-dimethylbutane: $CH_3-CH(CH_3)-CH(CH_3)-CH_3$
There are a total of $5$ structural isomers.
$1$. $n$-hexane: $CH_3-CH_2-CH_2-CH_2-CH_2-CH_3$
$2$. $2$-methylpentane: $CH_3-CH(CH_3)-CH_2-CH_2-CH_3$
$3$. $3$-methylpentane: $CH_3-CH_2-CH(CH_3)-CH_2-CH_3$
$4$. $2,2$-dimethylbutane: $CH_3-C(CH_3)_2-CH_2-CH_3$
$5$. $2,3$-dimethylbutane: $CH_3-CH(CH_3)-CH(CH_3)-CH_3$
There are a total of $5$ structural isomers.
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198
ChemistryMediumMCQMHT CET · 2026
Which of the following molecules has a chiral carbon atom?
A
B
C
D
Solution
(B) chiral carbon atom is a carbon atom bonded to four different groups or atoms. Let us analyze each option:
$(A)$ The central carbon is bonded to $-H$, $-H$, $-Cl$, and $-Br$. Since two groups $(-H)$ are identical, it is not chiral.
$(B)$ The central carbon is bonded to $-H$, $-CH_3$, $-OH$, and $-I$. All four groups are different, so this carbon is chiral.
$(C)$ The central carbon is bonded to $-CH_3$, $-CH_3$, $-Cl$, and $-Br$. Since two groups $(-CH_3)$ are identical, it is not chiral.
$(D)$ The central carbon is bonded to $-H$, $-Cl$, $-C_2H_5$, and $-C_2H_5$. Since two groups $(-C_2H_5)$ are identical, it is not chiral.
Therefore, the molecule in option $(B)$ contains a chiral carbon atom.
$(A)$ The central carbon is bonded to $-H$, $-H$, $-Cl$, and $-Br$. Since two groups $(-H)$ are identical, it is not chiral.
$(B)$ The central carbon is bonded to $-H$, $-CH_3$, $-OH$, and $-I$. All four groups are different, so this carbon is chiral.
$(C)$ The central carbon is bonded to $-CH_3$, $-CH_3$, $-Cl$, and $-Br$. Since two groups $(-CH_3)$ are identical, it is not chiral.
$(D)$ The central carbon is bonded to $-H$, $-Cl$, $-C_2H_5$, and $-C_2H_5$. Since two groups $(-C_2H_5)$ are identical, it is not chiral.
Therefore, the molecule in option $(B)$ contains a chiral carbon atom.
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199
ChemistryMediumMCQMHT CET · 2026
Which of the following has the highest nucleophilicity?
A
$F^-$
B
$OH^-$
C
$CH_3^-$
D
$NH_2^-$
Solution
(C) Nucleophilicity is related to the ability of a species to donate an electron pair. In a given period of the periodic table, nucleophilicity decreases as electronegativity increases.
Comparing the atoms $C$, $N$, $O$, and $F$ in the second period:
Electronegativity order: $C < N < O < F$.
Therefore, the basicity and nucleophilicity order is: $CH_3^- > NH_2^- > OH^- > F^-$.
Thus, $CH_3^-$ is the strongest nucleophile among the given options.
Comparing the atoms $C$, $N$, $O$, and $F$ in the second period:
Electronegativity order: $C < N < O < F$.
Therefore, the basicity and nucleophilicity order is: $CH_3^- > NH_2^- > OH^- > F^-$.
Thus, $CH_3^-$ is the strongest nucleophile among the given options.
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200
ChemistryMediumMCQMHT CET · 2026
Which one of the following series contains only electrophiles?
A
$R-OH, Cl^+, NH_3$
B
$H_2O, R_2N^-, H_3O^+$
C
$BF_3, SO_3, NO^+$
D
$AlCl_3, R-NH_2, PCl_3$
Solution
(C) $1$. An electrophile is an electron-deficient species that can accept an electron pair.
$2$. In option $C$, $BF_3$ is an electron-deficient molecule (octet incomplete), $SO_3$ has an electron-deficient sulfur atom due to electronegative oxygen atoms, and $NO^+$ is a positively charged ion.
$3$. All three species in option $C$ act as electrophiles.
$4$. Other options contain nucleophiles like $NH_3$, $R_2N^-$, $R-NH_2$, and $PCl_3$ (which has a lone pair).
$2$. In option $C$, $BF_3$ is an electron-deficient molecule (octet incomplete), $SO_3$ has an electron-deficient sulfur atom due to electronegative oxygen atoms, and $NO^+$ is a positively charged ion.
$3$. All three species in option $C$ act as electrophiles.
$4$. Other options contain nucleophiles like $NH_3$, $R_2N^-$, $R-NH_2$, and $PCl_3$ (which has a lone pair).
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201
ChemistryMediumMCQMHT CET · 2026
Which of the following colligative properties is useful to determine the molar masses of proteins, polymers, or colloids with the greatest precision?
A
Elevation in boiling point
B
Depression in freezing point
C
Osmotic pressure
D
Relative lowering of vapour pressure
Solution
(C) Step $1$: The molar mass of macromolecules like proteins and polymers is very high.
Step $2$: For these substances, the values of other colligative properties like elevation in boiling point or depression in freezing point are too small to be measured accurately.
Step $3$: Osmotic pressure $(\pi)$ is given by the equation $\pi = CRT = \frac{n}{V}RT = \frac{w}{MV}RT$. Even for dilute solutions of macromolecules, the osmotic pressure value is large enough to be measured with high precision.
Step $4$: Therefore, osmotic pressure is the most suitable method for determining the molar masses of such substances.
Step $2$: For these substances, the values of other colligative properties like elevation in boiling point or depression in freezing point are too small to be measured accurately.
Step $3$: Osmotic pressure $(\pi)$ is given by the equation $\pi = CRT = \frac{n}{V}RT = \frac{w}{MV}RT$. Even for dilute solutions of macromolecules, the osmotic pressure value is large enough to be measured with high precision.
Step $4$: Therefore, osmotic pressure is the most suitable method for determining the molar masses of such substances.
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202
ChemistryDifficultMCQMHT CET · 2026
If a $0.4 \text{ molal}$ solution of a nonvolatile solute in an organic solvent decreases its freezing point by $2.4 \text{ K}$, then the molal depression constant $(K_f)$ of the solvent will be:
A
$3.0 \text{ K kg mol}^{-1}$
B
$4.0 \text{ K kg mol}^{-1}$
C
$5.0 \text{ K kg mol}^{-1}$
D
$6.0 \text{ K kg mol}^{-1}$
Solution
(D) The formula for freezing point depression is $\Delta T_f = K_f \times m$, where $\Delta T_f$ is the depression in freezing point, $K_f$ is the molal depression constant, and $m$ is the molality of the solution.
Given:
$\Delta T_f = 2.4 \text{ K}$
$m = 0.4 \text{ mol kg}^{-1}$
Substituting the values into the formula:
$2.4 \text{ K} = K_f \times 0.4 \text{ mol kg}^{-1}$
$K_f = \frac{2.4 \text{ K}}{0.4 \text{ mol kg}^{-1}}$
$K_f = 6.0 \text{ K kg mol}^{-1}$
Therefore, the correct option is $D$.
Given:
$\Delta T_f = 2.4 \text{ K}$
$m = 0.4 \text{ mol kg}^{-1}$
Substituting the values into the formula:
$2.4 \text{ K} = K_f \times 0.4 \text{ mol kg}^{-1}$
$K_f = \frac{2.4 \text{ K}}{0.4 \text{ mol kg}^{-1}}$
$K_f = 6.0 \text{ K kg mol}^{-1}$
Therefore, the correct option is $D$.
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203
ChemistryEasyMCQMHT CET · 2026
Which of the following is a colligative property of a solution?
A
Vapour pressure
B
Boiling point
C
Osmotic pressure
D
Freezing point
Solution
(C) $1$. Colligative properties are properties of solutions that depend only on the number of solute particles present in a given amount of solvent, not on their identity.
$2$. Common colligative properties include relative lowering of vapour pressure, elevation of boiling point, depression of freezing point, and osmotic pressure.
$3$. Among the given options, $osmotic \ pressure$ is a colligative property, whereas vapour pressure, boiling point, and freezing point are properties that change due to the presence of solute but are not colligative properties themselves (their changes are).
$2$. Common colligative properties include relative lowering of vapour pressure, elevation of boiling point, depression of freezing point, and osmotic pressure.
$3$. Among the given options, $osmotic \ pressure$ is a colligative property, whereas vapour pressure, boiling point, and freezing point are properties that change due to the presence of solute but are not colligative properties themselves (their changes are).
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204
ChemistryDifficultMCQMHT CET · 2026
Calculate the boiling point of an aqueous solution containing $18 \text{ g}$ of glucose in $100 \text{ g}$ of water, given that the molal elevation constant $(K_b)$ of water is $0.5 \text{ K kg mol}^{-1}$. [Molar mass of glucose $= 180 \text{ g mol}^{-1}$ and boiling point of pure water $= 100^{\circ}\text{C}$] (in $^{\circ}\text{C}$)
A
$100$
B
$100.5$
C
$101.0$
D
$101.5$
Solution
(B) Step $1$: Calculate the number of moles of glucose $(n)$: $n = \frac{\text{mass}}{\text{molar mass}} = \frac{18 \text{ g}}{180 \text{ g mol}^{-1}} = 0.1 \text{ mol}$.
Step $2$: Calculate the molality $(m)$ of the solution: $m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.1 \text{ mol}}{0.1 \text{ kg}} = 1 \text{ mol kg}^{-1}$.
Step $3$: Calculate the elevation in boiling point $(\Delta T_b)$: $\Delta T_b = K_b \times m = 0.5 \text{ K kg mol}^{-1} \times 1 \text{ mol kg}^{-1} = 0.5 \text{ K}$ (or $0.5^{\circ}\text{C}$).
Step $4$: Calculate the boiling point of the solution: $T_b = T_b^{\circ} + \Delta T_b = 100^{\circ}\text{C} + 0.5^{\circ}\text{C} = 100.5^{\circ}\text{C}$.
Step $2$: Calculate the molality $(m)$ of the solution: $m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.1 \text{ mol}}{0.1 \text{ kg}} = 1 \text{ mol kg}^{-1}$.
Step $3$: Calculate the elevation in boiling point $(\Delta T_b)$: $\Delta T_b = K_b \times m = 0.5 \text{ K kg mol}^{-1} \times 1 \text{ mol kg}^{-1} = 0.5 \text{ K}$ (or $0.5^{\circ}\text{C}$).
Step $4$: Calculate the boiling point of the solution: $T_b = T_b^{\circ} + \Delta T_b = 100^{\circ}\text{C} + 0.5^{\circ}\text{C} = 100.5^{\circ}\text{C}$.
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205
ChemistryDifficultMCQMHT CET · 2026
$18 \text{ g}$ of glucose (molar mass $= 180 \text{ g/mol}$) is dissolved in water to prepare $500 \text{ ml}$ solution at $15^{\circ}\text{C}$. Calculate the osmotic pressure of the solution. $[R = 0.0821 \text{ L atm K}^{-1} \text{ mol}^{-1}]$ (in $\text{ atm}$)
A
$1.65$
B
$4.73$
C
$5.57$
D
$2.34$
Solution
(B) Step $1$: Calculate the number of moles of glucose $(n)$: $n = \frac{\text{mass}}{\text{molar mass}} = \frac{18 \text{ g}}{180 \text{ g/mol}} = 0.1 \text{ mol}$.
Step $2$: Convert volume to liters $(V)$: $V = 500 \text{ ml} = 0.5 \text{ L}$.
Step $3$: Convert temperature to Kelvin $(T)$: $T = 15 + 273.15 = 288.15 \text{ K}$.
Step $4$: Use the osmotic pressure formula $\pi = CRT = \frac{n}{V}RT$: $\pi = \frac{0.1 \text{ mol}}{0.5 \text{ L}} \times 0.0821 \text{ L atm K}^{-1} \text{ mol}^{-1} \times 288.15 \text{ K} = 0.2 \times 0.0821 \times 288.15 = 4.732 \text{ atm}$.
Thus, the osmotic pressure is approximately $4.73 \text{ atm}$.
Step $2$: Convert volume to liters $(V)$: $V = 500 \text{ ml} = 0.5 \text{ L}$.
Step $3$: Convert temperature to Kelvin $(T)$: $T = 15 + 273.15 = 288.15 \text{ K}$.
Step $4$: Use the osmotic pressure formula $\pi = CRT = \frac{n}{V}RT$: $\pi = \frac{0.1 \text{ mol}}{0.5 \text{ L}} \times 0.0821 \text{ L atm K}^{-1} \text{ mol}^{-1} \times 288.15 \text{ K} = 0.2 \times 0.0821 \times 288.15 = 4.732 \text{ atm}$.
Thus, the osmotic pressure is approximately $4.73 \text{ atm}$.
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206
ChemistryDifficultMCQMHT CET · 2026
The normal boiling point of ethyl acetate is $77^{\circ}\text{C}$. $A$ solution of a non-volatile, non-electrolyte solute in ethyl acetate boils at $78^{\circ}\text{C}$. The $K_b$ for ethyl acetate is $2.77^{\circ}\text{C kg mol}^{-1}$. What is the molality of the solution (in $\text{ m}$)?
A
$0.361$
B
$0.052$
C
$0.075$
D
$0.25$
Solution
(A) Step $1$: Calculate the elevation in boiling point $(\Delta T_b)$.
$\Delta T_b = T_b - T_b^{\circ} = 78^{\circ}\text{C} - 77^{\circ}\text{C} = 1^{\circ}\text{C}$.
Step $2$: Use the formula for elevation in boiling point: $\Delta T_b = K_b \times m$.
Step $3$: Rearrange to solve for molality $(m)$: $m = \frac{\Delta T_b}{K_b}$.
Step $4$: Substitute the values: $m = \frac{1}{2.77} \approx 0.361\text{ m}$.
$\Delta T_b = T_b - T_b^{\circ} = 78^{\circ}\text{C} - 77^{\circ}\text{C} = 1^{\circ}\text{C}$.
Step $2$: Use the formula for elevation in boiling point: $\Delta T_b = K_b \times m$.
Step $3$: Rearrange to solve for molality $(m)$: $m = \frac{\Delta T_b}{K_b}$.
Step $4$: Substitute the values: $m = \frac{1}{2.77} \approx 0.361\text{ m}$.
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207
ChemistryDifficultMCQMHT CET · 2026
Calculate the molar mass of a non-volatile solute when $4 \text{ g}$ of it is dissolved in $100 \text{ g}$ of a solvent that boils at $319.4 \text{ K}$. Given: $K_b = 2.4 \text{ K kg mol}^{-1}$ and the boiling point of the pure solvent is $319 \text{ K}$.
A
$220 \text{ g mol}^{-1}$
B
$230 \text{ g mol}^{-1}$
C
$240 \text{ g mol}^{-1}$
D
$250 \text{ g mol}^{-1}$
Solution
(C) Step $1$: Calculate the elevation in boiling point $\Delta T_b = T_b - T_b^\circ = 319.4 \text{ K} - 319 \text{ K} = 0.4 \text{ K}$.
Step $2$: Use the formula $\Delta T_b = K_b \times m$, where $m$ is molality.
Step $3$: Molality $m = \frac{w_2 \times 1000}{M_2 \times w_1}$, where $w_2 = 4 \text{ g}$, $w_1 = 100 \text{ g}$, and $M_2$ is the molar mass of the solute.
Step $4$: Substitute the values: $0.4 = 2.4 \times \frac{4 \times 1000}{M_2 \times 100}$.
Step $5$: Simplify: $0.4 = 2.4 \times \frac{40}{M_2} \implies 0.4 = \frac{96}{M_2}$.
Step $6$: Solve for $M_2$: $M_2 = \frac{96}{0.4} = 240 \text{ g mol}^{-1}$.
Step $2$: Use the formula $\Delta T_b = K_b \times m$, where $m$ is molality.
Step $3$: Molality $m = \frac{w_2 \times 1000}{M_2 \times w_1}$, where $w_2 = 4 \text{ g}$, $w_1 = 100 \text{ g}$, and $M_2$ is the molar mass of the solute.
Step $4$: Substitute the values: $0.4 = 2.4 \times \frac{4 \times 1000}{M_2 \times 100}$.
Step $5$: Simplify: $0.4 = 2.4 \times \frac{40}{M_2} \implies 0.4 = \frac{96}{M_2}$.
Step $6$: Solve for $M_2$: $M_2 = \frac{96}{0.4} = 240 \text{ g mol}^{-1}$.
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208
ChemistryDifficultMCQMHT CET · 2026
Calculate the van't Hoff factor $(i)$ for an aqueous solution of $0.02 \text{ m}$ formic acid if it freezes at $-0.045^{\circ}\text{C}$. Given: $K_f = 1.86 \text{ K kg mol}^{-1}$ and the freezing point of pure water $= 0^{\circ}\text{C}$.
A
$1.21$
B
$1.46$
C
$1.68$
D
$1.05$
Solution
(A) Step $1$: Calculate the depression in freezing point $(\Delta T_f)$.
$\Delta T_f = T_f^{\circ} - T_f = 0^{\circ}\text{C} - (-0.045^{\circ}\text{C}) = 0.045 \text{ K}$.
Step $2$: Use the formula for depression in freezing point: $\Delta T_f = i \times K_f \times m$.
Step $3$: Substitute the given values: $0.045 = i \times 1.86 \times 0.02$.
Step $4$: Solve for $i$: $i = \frac{0.045}{1.86 \times 0.02} = \frac{0.045}{0.0372} \approx 1.21$.
$\Delta T_f = T_f^{\circ} - T_f = 0^{\circ}\text{C} - (-0.045^{\circ}\text{C}) = 0.045 \text{ K}$.
Step $2$: Use the formula for depression in freezing point: $\Delta T_f = i \times K_f \times m$.
Step $3$: Substitute the given values: $0.045 = i \times 1.86 \times 0.02$.
Step $4$: Solve for $i$: $i = \frac{0.045}{1.86 \times 0.02} = \frac{0.045}{0.0372} \approx 1.21$.
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209
ChemistryMediumMCQMHT CET · 2026
Which of the following concentrations of solutions of urea in water exhibits minimum freezing point depression (in $\text{ m}$)?
A
$0.12$
B
$0.18$
C
$0.08$
D
$0.06$
Solution
(D) The freezing point depression $\Delta T_f$ is given by the formula $\Delta T_f = i \cdot K_f \cdot m$.
For urea, which is a non-electrolyte, the van't Hoff factor $i = 1$.
Thus, $\Delta T_f = K_f \cdot m$.
Since $K_f$ is a constant for the solvent (water), $\Delta T_f$ is directly proportional to the molality $m$ of the solution.
To obtain the minimum freezing point depression, we must choose the solution with the minimum molality.
Comparing the given concentrations: $0.12 \text{ m}$, $0.18 \text{ m}$, $0.08 \text{ m}$, and $0.06 \text{ m}$, the smallest value is $0.06 \text{ m}$.
Therefore, the solution with $0.06 \text{ m}$ concentration exhibits the minimum freezing point depression.
For urea, which is a non-electrolyte, the van't Hoff factor $i = 1$.
Thus, $\Delta T_f = K_f \cdot m$.
Since $K_f$ is a constant for the solvent (water), $\Delta T_f$ is directly proportional to the molality $m$ of the solution.
To obtain the minimum freezing point depression, we must choose the solution with the minimum molality.
Comparing the given concentrations: $0.12 \text{ m}$, $0.18 \text{ m}$, $0.08 \text{ m}$, and $0.06 \text{ m}$, the smallest value is $0.06 \text{ m}$.
Therefore, the solution with $0.06 \text{ m}$ concentration exhibits the minimum freezing point depression.
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210
ChemistryMediumMCQMHT CET · 2026
Identify the correct statement from the following.
A
The vapour pressure of a solvent increases by dissolving a non-volatile solute into it.
B
The boiling point of a solvent decreases by dissolving a non-volatile solute into it.
C
The osmotic pressure of an electrolytic solution is greater than a non-electrolytic solution of the same concentration.
D
The freezing point of a solvent is a colligative property.
Solution
(C) $1$. According to Raoult's law, adding a non-volatile solute decreases the vapour pressure of the solvent.
$2$. The decrease in vapour pressure leads to an elevation in the boiling point of the solvent.
$3$. Osmotic pressure is a colligative property, which depends on the number of particles. Electrolytes dissociate into multiple ions, increasing the number of particles compared to non-electrolytes at the same concentration, thus resulting in higher osmotic pressure.
$4$. Freezing point depression is a colligative property, but the freezing point itself is not.
$2$. The decrease in vapour pressure leads to an elevation in the boiling point of the solvent.
$3$. Osmotic pressure is a colligative property, which depends on the number of particles. Electrolytes dissociate into multiple ions, increasing the number of particles compared to non-electrolytes at the same concentration, thus resulting in higher osmotic pressure.
$4$. Freezing point depression is a colligative property, but the freezing point itself is not.
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211
ChemistryDifficultMCQMHT CET · 2026
Calculate the cryoscopic constant $(K_f)$ of a solvent if the depression in freezing point of a $0.3 \text{ m}$ solution of a nonelectrolyte is $0.48 \text{ K}$.
A
$1.1 \text{ K kg mol}^{-1}$
B
$1.6 \text{ K kg mol}^{-1}$
C
$2.2 \text{ K kg mol}^{-1}$
D
$2.6 \text{ K kg mol}^{-1}$
Solution
(B) The formula for depression in freezing point is: $\Delta T_f = K_f \times m$
Given:
$\Delta T_f = 0.48 \text{ K}$
$m = 0.3 \text{ mol kg}^{-1}$
Rearranging the formula to solve for $K_f$:
$K_f = \frac{\Delta T_f}{m}$
$K_f = \frac{0.48 \text{ K}}{0.3 \text{ mol kg}^{-1}}$
$K_f = 1.6 \text{ K kg mol}^{-1}$
Thus, the correct option is $B$.
Given:
$\Delta T_f = 0.48 \text{ K}$
$m = 0.3 \text{ mol kg}^{-1}$
Rearranging the formula to solve for $K_f$:
$K_f = \frac{\Delta T_f}{m}$
$K_f = \frac{0.48 \text{ K}}{0.3 \text{ mol kg}^{-1}}$
$K_f = 1.6 \text{ K kg mol}^{-1}$
Thus, the correct option is $B$.
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212
ChemistryDifficultMCQMHT CET · 2026
Calculate the molar mass of a nonvolatile solute if $6.4 \text{ g}$ of it dissolved in $100 \text{ g}$ water produces a relative lowering in vapour pressure of $0.016$ at $300 \text{ K}$.
A
$60 \text{ g mol}^{-1}$
B
$66 \text{ g mol}^{-1}$
C
$72 \text{ g mol}^{-1}$
D
$84 \text{ g mol}^{-1}$
Solution
(C) The relative lowering in vapour pressure is given by Raoult's Law: $\frac{P^\circ - P_s}{P^\circ} = \frac{n_2}{n_1 + n_2} \approx \frac{n_2}{n_1}$ for dilute solutions.
Given: $\frac{P^\circ - P_s}{P^\circ} = 0.016$, mass of solute $w_2 = 6.4 \text{ g}$, mass of solvent $w_1 = 100 \text{ g}$, molar mass of water $M_1 = 18 \text{ g mol}^{-1}$.
Formula: $\frac{P^\circ - P_s}{P^\circ} = \frac{w_2 \times M_1}{M_2 \times w_1}$.
Substitute the values: $0.016 = \frac{6.4 \times 18}{M_2 \times 100}$.
$M_2 = \frac{6.4 \times 18}{0.016 \times 100} = \frac{115.2}{1.6} = 72 \text{ g mol}^{-1}$.
Given: $\frac{P^\circ - P_s}{P^\circ} = 0.016$, mass of solute $w_2 = 6.4 \text{ g}$, mass of solvent $w_1 = 100 \text{ g}$, molar mass of water $M_1 = 18 \text{ g mol}^{-1}$.
Formula: $\frac{P^\circ - P_s}{P^\circ} = \frac{w_2 \times M_1}{M_2 \times w_1}$.
Substitute the values: $0.016 = \frac{6.4 \times 18}{M_2 \times 100}$.
$M_2 = \frac{6.4 \times 18}{0.016 \times 100} = \frac{115.2}{1.6} = 72 \text{ g mol}^{-1}$.
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213
ChemistryMediumMCQMHT CET · 2026
Identify the correct statement regarding $0.2 \text{ M}$ urea and $0.2 \text{ M}$ sucrose solutions.
A
The urea solution is hypertonic to the sucrose solution.
B
The osmotic pressure of the urea solution is higher than that of the sucrose solution.
C
These solutions exhibit the same osmotic pressure.
D
The sucrose solution is hypotonic to the urea solution.
Solution
(C) Step $1$: Osmotic pressure $(\pi)$ is given by the formula $\pi = iCRT$, where $i$ is the van't Hoff factor, $C$ is molarity, $R$ is the gas constant, and $T$ is temperature.
Step $2$: For urea $(NH_2CONH_2)$, $i = 1$ (non-electrolyte). For sucrose $(C_{12}H_{22}O_{11})$, $i = 1$ (non-electrolyte).
Step $3$: Both solutions have the same molarity $(C = 0.2 \text{ M})$. Since $i$, $C$, $R$, and $T$ are identical for both, their osmotic pressures are equal.
Step $4$: Solutions with the same osmotic pressure are called isotonic solutions. Thus, statement $(c)$ is correct.
Step $2$: For urea $(NH_2CONH_2)$, $i = 1$ (non-electrolyte). For sucrose $(C_{12}H_{22}O_{11})$, $i = 1$ (non-electrolyte).
Step $3$: Both solutions have the same molarity $(C = 0.2 \text{ M})$. Since $i$, $C$, $R$, and $T$ are identical for both, their osmotic pressures are equal.
Step $4$: Solutions with the same osmotic pressure are called isotonic solutions. Thus, statement $(c)$ is correct.
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214
ChemistryDifficultMCQMHT CET · 2026
Calculate the molality of a solution of a non-volatile solute if the boiling point of the solution and the molal elevation constant for the solvent are $319.8 \text{ K}$ and $2.5 \text{ K kg mol}^{-1}$ respectively. [Boiling point of pure solvent $= 319.5 \text{ K}$]
A
$0.12 \text{ mol kg}^{-1}$
B
$0.14 \text{ mol kg}^{-1}$
C
$0.17 \text{ mol kg}^{-1}$
D
$0.21 \text{ mol kg}^{-1}$
Solution
(A) Step $1$: Calculate the elevation in boiling point $(\Delta T_b)$.
$\Delta T_b = T_b - T_b^\circ = 319.8 \text{ K} - 319.5 \text{ K} = 0.3 \text{ K}$.
Step $2$: Use the formula for elevation in boiling point: $\Delta T_b = K_b \times m$.
Step $3$: Rearrange to solve for molality $(m)$:
$m = \frac{\Delta T_b}{K_b} = \frac{0.3 \text{ K}}{2.5 \text{ K kg mol}^{-1}} = 0.12 \text{ mol kg}^{-1}$.
$\Delta T_b = T_b - T_b^\circ = 319.8 \text{ K} - 319.5 \text{ K} = 0.3 \text{ K}$.
Step $2$: Use the formula for elevation in boiling point: $\Delta T_b = K_b \times m$.
Step $3$: Rearrange to solve for molality $(m)$:
$m = \frac{\Delta T_b}{K_b} = \frac{0.3 \text{ K}}{2.5 \text{ K kg mol}^{-1}} = 0.12 \text{ mol kg}^{-1}$.
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215
ChemistryMediumMCQMHT CET · 2026
Which of the following options is true when an electrolyte solution is diluted?
A
both $\Lambda_m$ and $\kappa$ increase
B
both $\Lambda_m$ and $\kappa$ decrease
C
$\Lambda_m$ increases and $\kappa$ decreases
D
$\Lambda_m$ decreases and $\kappa$ increases
Solution
(C) $1$. Molar conductivity $(\Lambda_m)$ is defined as $\Lambda_m = \frac{\kappa}{C}$. As the solution is diluted, the concentration $(C)$ decreases, which leads to an increase in the number of ions per unit volume available for conduction, causing $\Lambda_m$ to increase.
$2$. Conductivity $(\kappa)$ is defined as the conductance of a solution contained between two electrodes of unit area and unit distance apart. Upon dilution, the number of ions per unit volume decreases, which leads to a decrease in conductivity $(\kappa)$.
$3$. Therefore, upon dilution, $\Lambda_m$ increases and $\kappa$ decreases.
$2$. Conductivity $(\kappa)$ is defined as the conductance of a solution contained between two electrodes of unit area and unit distance apart. Upon dilution, the number of ions per unit volume decreases, which leads to a decrease in conductivity $(\kappa)$.
$3$. Therefore, upon dilution, $\Lambda_m$ increases and $\kappa$ decreases.
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216
ChemistryEasyMCQMHT CET · 2026
Identify the correct name of the law: "At infinite dilution, each ion migrates independently of its co-ion and contributes to the total molar conductivity of an electrolyte, irrespective of the nature of the other ion to which it is associated."
A
Henry's law
B
Raoult's law
C
Nernst derivative law
D
Kohlrausch's law of independent migration of ions
Solution
(D) Step $1$: The statement describes the behavior of ions in an electrolytic solution at infinite dilution.
Step $2$: According to Kohlrausch's law of independent migration of ions, at infinite dilution, where dissociation is complete, each ion makes a definite contribution to the molar conductivity of the electrolyte, which is independent of the presence of other ions.
Step $3$: Therefore, the correct law is Kohlrausch's law.
Step $2$: According to Kohlrausch's law of independent migration of ions, at infinite dilution, where dissociation is complete, each ion makes a definite contribution to the molar conductivity of the electrolyte, which is independent of the presence of other ions.
Step $3$: Therefore, the correct law is Kohlrausch's law.
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217
ChemistryDifficultMCQMHT CET · 2026
Calculate the molar conductivity of $0.2 \text{ M}$ $KCl$ solution at $298 \text{ K}$ if the conductivity $(k)$ is $0.0248 \text{ } \Omega^{-1} \text{cm}^{-1}$.
A
$143 \text{ } \Omega^{-1} \text{cm}^2 \text{mol}^{-1}$
B
$98 \text{ } \Omega^{-1} \text{cm}^2 \text{mol}^{-1}$
C
$124 \text{ } \Omega^{-1} \text{cm}^2 \text{mol}^{-1}$
D
$87 \text{ } \Omega^{-1} \text{cm}^2 \text{mol}^{-1}$
Solution
(C) The formula for molar conductivity $(\Lambda_m)$ is: $\Lambda_m = \frac{1000 \times k}{M}$
Given: Conductivity $(k)$ = $0.0248 \text{ } \Omega^{-1} \text{cm}^{-1}$, Molarity $(M)$ = $0.2 \text{ M}$.
Substituting the values: $\Lambda_m = \frac{1000 \times 0.0248}{0.2}$
$\Lambda_m = \frac{24.8}{0.2} = 124 \text{ } \Omega^{-1} \text{cm}^2 \text{mol}^{-1}$.
Given: Conductivity $(k)$ = $0.0248 \text{ } \Omega^{-1} \text{cm}^{-1}$, Molarity $(M)$ = $0.2 \text{ M}$.
Substituting the values: $\Lambda_m = \frac{1000 \times 0.0248}{0.2}$
$\Lambda_m = \frac{24.8}{0.2} = 124 \text{ } \Omega^{-1} \text{cm}^2 \text{mol}^{-1}$.
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218
ChemistryDifficultMCQMHT CET · 2026
The molar conductivity of a $0.04 \text{ M}$ $AB_2$ type salt solution at $300 \text{ K}$ is $200 \text{ } \Omega^{-1} \text{ cm}^2 \text{ mol}^{-1}$. Find the conductivity.
A
$0.006 \text{ } \Omega^{-1} \text{ cm}^{-1}$
B
$0.008 \text{ } \Omega^{-1} \text{ cm}^{-1}$
C
$0.01 \text{ } \Omega^{-1} \text{ cm}^{-1}$
D
$0.015 \text{ } \Omega^{-1} \text{ cm}^{-1}$
Solution
(B) The relationship between molar conductivity $(\Lambda_m)$ and conductivity $(\kappa)$ is given by the formula: $\Lambda_m = \frac{\kappa \times 1000}{M}$.
Rearranging for conductivity $(\kappa)$: $\kappa = \frac{\Lambda_m \times M}{1000}$.
Given: $\Lambda_m = 200 \text{ } \Omega^{-1} \text{ cm}^2 \text{ mol}^{-1}$, $M = 0.04 \text{ M}$.
Substituting the values: $\kappa = \frac{200 \times 0.04}{1000}$.
$\kappa = \frac{8}{1000} = 0.008 \text{ } \Omega^{-1} \text{ cm}^{-1}$.
Rearranging for conductivity $(\kappa)$: $\kappa = \frac{\Lambda_m \times M}{1000}$.
Given: $\Lambda_m = 200 \text{ } \Omega^{-1} \text{ cm}^2 \text{ mol}^{-1}$, $M = 0.04 \text{ M}$.
Substituting the values: $\kappa = \frac{200 \times 0.04}{1000}$.
$\kappa = \frac{8}{1000} = 0.008 \text{ } \Omega^{-1} \text{ cm}^{-1}$.
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219
ChemistryDifficultMCQMHT CET · 2026
The molar conductivity of a $0.01 \text{ M}$ monobasic acid at $25^{\circ} \text{C}$ is $15 \text{ } \Omega^{-1} \text{cm}^2 \text{mol}^{-1}$. The molar conductivity of the same acid at infinite dilution is $375 \text{ } \Omega^{-1} \text{cm}^2 \text{mol}^{-1}$. Calculate the concentration of $[H^+]$ in the solution.
A
$2.0 \times 10^{-4} \text{ M}$
B
$3.0 \times 10^{-4} \text{ M}$
C
$4.0 \times 10^{-4} \text{ M}$
D
$5.0 \times 10^{-4} \text{ M}$
Solution
(C) Step $1$: Calculate the degree of dissociation $(\alpha)$ using the formula $\alpha = \frac{\Lambda_m}{\Lambda_m^\circ}$.
$\alpha = \frac{15}{375} = 0.04$.
Step $2$: Calculate the concentration of $[H^+]$ using the relation $[H^+] = C \times \alpha$.
Given $C = 0.01 \text{ M}$.
$[H^+] = 0.01 \times 0.04 = 4 \times 10^{-4} \text{ M}$.
$\alpha = \frac{15}{375} = 0.04$.
Step $2$: Calculate the concentration of $[H^+]$ using the relation $[H^+] = C \times \alpha$.
Given $C = 0.01 \text{ M}$.
$[H^+] = 0.01 \times 0.04 = 4 \times 10^{-4} \text{ M}$.
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220
ChemistryDifficultMCQMHT CET · 2026
Calculate the concentration of silver nitrate solution if molar conductivity and conductivity of silver nitrate at $25^{\circ} \text{C}$ are respectively $120 \text{ } \Omega^{-1} \text{cm}^2 \text{mol}^{-1}$ and $0.0024 \text{ } \Omega^{-1} \text{cm}^{-1}$. (in $\text{ M}$)
A
$0.01$
B
$0.02$
C
$0.03$
D
$0.04$
Solution
(B) The relationship between molar conductivity $(\Lambda_m)$, conductivity $(k)$, and molarity $(M)$ is given by: $\Lambda_m = \frac{1000 \times k}{M}$.
Rearranging for molarity: $M = \frac{1000 \times k}{\Lambda_m}$.
Given: $k = 0.0024 \text{ } \Omega^{-1} \text{cm}^{-1}$ and $\Lambda_m = 120 \text{ } \Omega^{-1} \text{cm}^2 \text{mol}^{-1}$.
Substituting the values: $M = \frac{1000 \times 0.0024}{120}$.
$M = \frac{2.4}{120} = 0.02 \text{ M}$.
Rearranging for molarity: $M = \frac{1000 \times k}{\Lambda_m}$.
Given: $k = 0.0024 \text{ } \Omega^{-1} \text{cm}^{-1}$ and $\Lambda_m = 120 \text{ } \Omega^{-1} \text{cm}^2 \text{mol}^{-1}$.
Substituting the values: $M = \frac{1000 \times 0.0024}{120}$.
$M = \frac{2.4}{120} = 0.02 \text{ M}$.
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221
ChemistryMediumMCQMHT CET · 2026
The graphical variation of molar conductivity $(\Lambda_m)$ against the square root of molar concentration $(\sqrt{c})$ for a certain electrolyte '$X$' is linear with an intercept on the y-axis. Identify '$X$' from the following.
A
$CH_3COOH$
B
$NH_4OH$
C
$HCOOH$
D
$CH_3COONa$
Solution
(D) $1$. According to the Kohlrausch law and the Debye-$H$ückel-Onsager equation, strong electrolytes show a linear relationship between molar conductivity $(\Lambda_m)$ and $\sqrt{c}$ given by $\Lambda_m = \Lambda_m^\circ - A\sqrt{c}$, where $\Lambda_m^\circ$ is the intercept on the y-axis.
$2$. Weak electrolytes like $CH_3COOH$, $NH_4OH$, and $HCOOH$ show a non-linear curve that increases sharply as concentration approaches zero.
$3$. $CH_3COONa$ is a strong electrolyte, which dissociates completely in solution, thus following the linear relationship.
$4$. Therefore, '$X$' is $CH_3COONa$.
$2$. Weak electrolytes like $CH_3COOH$, $NH_4OH$, and $HCOOH$ show a non-linear curve that increases sharply as concentration approaches zero.
$3$. $CH_3COONa$ is a strong electrolyte, which dissociates completely in solution, thus following the linear relationship.
$4$. Therefore, '$X$' is $CH_3COONa$.
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222
ChemistryDifficultMCQMHT CET · 2026
The conductivity of $0.02 \text{ M}$ $KCl$ solution at $298 \text{ K}$ is $0.0123 \text{ } \Omega^{-1} \text{ cm}^{-1}$. If the resistance of the cell containing this solution is $120 \text{ } \Omega$, what is the value of the cell constant (in $\text{ cm}^{-1}$)?
A
$1.050$
B
$1.025$
C
$1.376$
D
$1.476$
Solution
(D) The cell constant $(G^*)$ is given by the formula: $G^* = \kappa \times R$
Given conductivity $(\kappa)$ = $0.0123 \text{ } \Omega^{-1} \text{ cm}^{-1}$
Given resistance $(R)$ = $120 \text{ } \Omega$
$G^* = 0.0123 \text{ } \Omega^{-1} \text{ cm}^{-1} \times 120 \text{ } \Omega$
$G^* = 1.476 \text{ cm}^{-1}$
Given conductivity $(\kappa)$ = $0.0123 \text{ } \Omega^{-1} \text{ cm}^{-1}$
Given resistance $(R)$ = $120 \text{ } \Omega$
$G^* = 0.0123 \text{ } \Omega^{-1} \text{ cm}^{-1} \times 120 \text{ } \Omega$
$G^* = 1.476 \text{ cm}^{-1}$
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223
ChemistryDifficultMCQMHT CET · 2026
What is the conductivity of $0.02 \text{ M}$ $AgNO_3$ solution having cell constant $1.2 \text{ cm}^{-1}$ and resistance $95.0 \text{ } \Omega$?
A
$1.263 \times 10^{-2} \text{ } \Omega^{-1} \text{cm}^{-1}$
B
$1.263 \text{ } \Omega^{-1} \text{cm}^{-1}$
C
$1.340 \text{ } \Omega^{-1} \text{cm}^{-1}$
D
$1.463 \text{ } \Omega^{-1} \text{cm}^{-1}$
Solution
(A) The formula for conductivity $(\kappa)$ is given by: $\kappa = \frac{G^*}{R}$
Given: Cell constant $(G^*)$ = $1.2 \text{ cm}^{-1}$, Resistance $(R)$ = $95.0 \text{ } \Omega$.
Substituting the values: $\kappa = \frac{1.2 \text{ cm}^{-1}}{95.0 \text{ } \Omega} = 0.01263 \text{ } \Omega^{-1} \text{cm}^{-1}$.
Therefore, $\kappa = 1.263 \times 10^{-2} \text{ } \Omega^{-1} \text{cm}^{-1}$.
Given: Cell constant $(G^*)$ = $1.2 \text{ cm}^{-1}$, Resistance $(R)$ = $95.0 \text{ } \Omega$.
Substituting the values: $\kappa = \frac{1.2 \text{ cm}^{-1}}{95.0 \text{ } \Omega} = 0.01263 \text{ } \Omega^{-1} \text{cm}^{-1}$.
Therefore, $\kappa = 1.263 \times 10^{-2} \text{ } \Omega^{-1} \text{cm}^{-1}$.
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224
ChemistryDifficultMCQMHT CET · 2026
What must be the molarity of $BaCl_2$ solution to have molar conductivity $240 \text{ } \Omega^{-1} \text{cm}^2 \text{mol}^{-1}$ and conductivity $0.012 \text{ } \Omega^{-1} \text{cm}^{-1}$ at $25^{\circ} \text{C}$ (in $\text{ M}$)?
A
$0.01$
B
$0.02$
C
$0.05$
D
$0.1$
Solution
(C) The relationship between molar conductivity $(\Lambda_m)$, conductivity $(\kappa)$, and molarity $(M)$ is given by the formula:
$\Lambda_m = \frac{1000 \times \kappa}{M}$
Rearranging the formula to solve for molarity $(M)$:
$M = \frac{1000 \times \kappa}{\Lambda_m}$
Substitute the given values: $\kappa = 0.012 \text{ } \Omega^{-1} \text{cm}^{-1}$ and $\Lambda_m = 240 \text{ } \Omega^{-1} \text{cm}^2 \text{mol}^{-1}$.
$M = \frac{1000 \times 0.012}{240}$
$M = \frac{12}{240} = 0.05 \text{ M}$
$\Lambda_m = \frac{1000 \times \kappa}{M}$
Rearranging the formula to solve for molarity $(M)$:
$M = \frac{1000 \times \kappa}{\Lambda_m}$
Substitute the given values: $\kappa = 0.012 \text{ } \Omega^{-1} \text{cm}^{-1}$ and $\Lambda_m = 240 \text{ } \Omega^{-1} \text{cm}^2 \text{mol}^{-1}$.
$M = \frac{1000 \times 0.012}{240}$
$M = \frac{12}{240} = 0.05 \text{ M}$
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225
ChemistryDifficultMCQMHT CET · 2026
What is the molar conductivity of $CH_3COOH$ at infinite dilution if the molar conductivities of $H_2SO_4$, $K_2SO_4$, and $CH_3COOK$ at infinite dilution are $x$, $y$, and $z$ $\Omega^{-1} \text{cm}^2 \text{mol}^{-1}$ respectively?
A
$\frac{(x - y)}{2} + z$
B
$(x - y + 2z)$
C
$(x + y - z)$
D
$\frac{(x - y)}{2} + 2z$
Solution
(A) According to Kohlrausch's law:
$\Lambda^0_m(CH_3COOH) = \Lambda^0_m(CH_3COOK) + \Lambda^0_m(H^+) - \Lambda^0_m(K^+)$
We know that $\Lambda^0_m(H^+) = \frac{1}{2} \Lambda^0_m(H_2SO_4) = \frac{x}{2}$ and $\Lambda^0_m(K^+) = \frac{1}{2} \Lambda^0_m(K_2SO_4) = \frac{y}{2}$.
Substituting these values:
$\Lambda^0_m(CH_3COOH) = z + \frac{x}{2} - \frac{y}{2} = \frac{(x - y)}{2} + z$.
$\Lambda^0_m(CH_3COOH) = \Lambda^0_m(CH_3COOK) + \Lambda^0_m(H^+) - \Lambda^0_m(K^+)$
We know that $\Lambda^0_m(H^+) = \frac{1}{2} \Lambda^0_m(H_2SO_4) = \frac{x}{2}$ and $\Lambda^0_m(K^+) = \frac{1}{2} \Lambda^0_m(K_2SO_4) = \frac{y}{2}$.
Substituting these values:
$\Lambda^0_m(CH_3COOH) = z + \frac{x}{2} - \frac{y}{2} = \frac{(x - y)}{2} + z$.
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226
ChemistryDifficultMCQMHT CET · 2026
What is the molar conductivity of $0.02 \text{ M}$ $KI$ solution if its conductivity is $4.37 \times 10^{-4} \text{ } \Omega^{-1} \text{cm}^{-1}$?
A
$8.74 \text{ } \Omega^{-1} \text{cm}^2 \text{mol}^{-1}$
B
$21.85 \text{ } \Omega^{-1} \text{cm}^2 \text{mol}^{-1}$
C
$43.70 \text{ } \Omega^{-1} \text{cm}^2 \text{mol}^{-1}$
D
$13.65 \text{ } \Omega^{-1} \text{cm}^2 \text{mol}^{-1}$
Solution
(B) The formula for molar conductivity $(\Lambda_m)$ is given by: $\Lambda_m = \frac{\kappa \times 1000}{C}$
Given conductivity $(\kappa)$ = $4.37 \times 10^{-4} \text{ } \Omega^{-1} \text{cm}^{-1}$
Given concentration $(C)$ = $0.02 \text{ M}$
Substituting the values: $\Lambda_m = \frac{4.37 \times 10^{-4} \times 1000}{0.02}$
$\Lambda_m = \frac{0.437}{0.02} = 21.85 \text{ } \Omega^{-1} \text{cm}^2 \text{mol}^{-1}$
Given conductivity $(\kappa)$ = $4.37 \times 10^{-4} \text{ } \Omega^{-1} \text{cm}^{-1}$
Given concentration $(C)$ = $0.02 \text{ M}$
Substituting the values: $\Lambda_m = \frac{4.37 \times 10^{-4} \times 1000}{0.02}$
$\Lambda_m = \frac{0.437}{0.02} = 21.85 \text{ } \Omega^{-1} \text{cm}^2 \text{mol}^{-1}$
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227
ChemistryEasyMCQMHT CET · 2026
Which of the following expressions indicates the correct relationship between molar conductivity of a strong electrolyte and its concentration $c$?
A
$\Lambda_m = \Lambda_m^0 + \sqrt{c}$
B
$\Lambda_m = \Lambda_m^0 - A\sqrt{c}$
C
$\Lambda_m = \Lambda_m^0 + A\sqrt{c}$
D
$\Lambda_m = \Lambda_m^0 - \sqrt{c}$
Solution
(B) For strong electrolytes, the molar conductivity $\Lambda_m$ varies with the concentration $c$ according to the Kohlrausch equation:
$\Lambda_m = \Lambda_m^0 - A\sqrt{c}$
where $\Lambda_m^0$ is the molar conductivity at infinite dilution and $A$ is a constant that depends on the nature of the solvent and temperature.
$\Lambda_m = \Lambda_m^0 - A\sqrt{c}$
where $\Lambda_m^0$ is the molar conductivity at infinite dilution and $A$ is a constant that depends on the nature of the solvent and temperature.
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228
ChemistryDifficultMCQMHT CET · 2026
If molar conductance at infinite dilution for $CH_3COO^-$ and $H^+$ ions are $50 \text{ S cm}^2 \text{mol}^{-1}$ and $350 \text{ S cm}^2 \text{mol}^{-1}$ respectively, and the molar conductivity of $5 \times 10^{-2} \text{ M}$ $CH_3COOH$ is $20 \text{ S cm}^2 \text{mol}^{-1}$, what is the hydrogen ion concentration in $\text{mol/dm}^3$ of $CH_3COOH$?
A
$5 \times 10^{-2}$
B
$5 \times 10^{-3}$
C
$2.5 \times 10^{-3}$
D
$2.5 \times 10^{-4}$
Solution
(C) Step $1$: Calculate the molar conductivity at infinite dilution $(\Lambda_m^0)$ for $CH_3COOH$ using Kohlrausch's law:
$\Lambda_m^0(CH_3COOH) = \lambda^0(CH_3COO^-) + \lambda^0(H^+) = 50 + 350 = 400 \text{ S cm}^2 \text{mol}^{-1}$.
Step $2$: Calculate the degree of dissociation $(\alpha)$:
$\alpha = \frac{\Lambda_m^c}{\Lambda_m^0} = \frac{20}{400} = 0.05$.
Step $3$: Calculate the hydrogen ion concentration $[H^+]$:
$[H^+] = C \times \alpha = (5 \times 10^{-2} \text{ M}) \times 0.05 = 2.5 \times 10^{-3} \text{ mol/dm}^3$.
$\Lambda_m^0(CH_3COOH) = \lambda^0(CH_3COO^-) + \lambda^0(H^+) = 50 + 350 = 400 \text{ S cm}^2 \text{mol}^{-1}$.
Step $2$: Calculate the degree of dissociation $(\alpha)$:
$\alpha = \frac{\Lambda_m^c}{\Lambda_m^0} = \frac{20}{400} = 0.05$.
Step $3$: Calculate the hydrogen ion concentration $[H^+]$:
$[H^+] = C \times \alpha = (5 \times 10^{-2} \text{ M}) \times 0.05 = 2.5 \times 10^{-3} \text{ mol/dm}^3$.
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229
ChemistryEasyMCQMHT CET · 2026
Which of the following statements is true about the voltaic cell?
A
It converts chemical energy into electrical energy.
B
It converts electrical energy into chemical energy.
C
The anode of voltaic cells is positive.
D
The cathode of voltaic cell is negative.
Solution
(A) Step $1$: $A$ voltaic cell (or galvanic cell) is an electrochemical cell that derives electrical energy from spontaneous redox reactions occurring within the cell.
Step $2$: In a voltaic cell, chemical energy is converted into electrical energy.
Step $3$: By convention, the anode is the electrode where oxidation occurs (negative terminal) and the cathode is the electrode where reduction occurs (positive terminal).
Step $4$: Therefore, option $A$ is the correct statement.
Step $2$: In a voltaic cell, chemical energy is converted into electrical energy.
Step $3$: By convention, the anode is the electrode where oxidation occurs (negative terminal) and the cathode is the electrode where reduction occurs (positive terminal).
Step $4$: Therefore, option $A$ is the correct statement.
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230
ChemistryMediumMCQMHT CET · 2026
Which of the following statements regarding a mercury battery is $NOT$ true?
A
It is a primary dry cell.
B
It consists of a $Zn$ anode amalgamated with mercury.
C
The electrolyte is strongly acidic.
D
In this, $Hg$ is obtained by the reduction of $HgO$.
Solution
(C) Step $1$: $A$ mercury battery is a primary cell that uses a $Zn$ anode and a $HgO$ cathode.
Step $2$: The electrolyte used is a paste of $KOH$ and $ZnO$, which is strongly alkaline (basic), not acidic.
Step $3$: The cell reaction at the cathode is $HgO(s) + H_2O(l) + 2e^- \rightarrow Hg(l) + 2OH^-(aq)$, where $HgO$ is reduced to $Hg$.
Step $4$: Since the electrolyte is basic, the statement that it is strongly acidic is false.
Step $2$: The electrolyte used is a paste of $KOH$ and $ZnO$, which is strongly alkaline (basic), not acidic.
Step $3$: The cell reaction at the cathode is $HgO(s) + H_2O(l) + 2e^- \rightarrow Hg(l) + 2OH^-(aq)$, where $HgO$ is reduced to $Hg$.
Step $4$: Since the electrolyte is basic, the statement that it is strongly acidic is false.
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231
ChemistryEasyMCQMHT CET · 2026
Which of the following is used as an electrolyte in a dry cell?
A
Potassium hydroxide
B
Sulphuric acid
C
Ammonium chloride and zinc chloride
D
Manganese dioxide
Solution
(C) In a dry cell (Leclanché cell), the electrolyte is a moist paste of $NH_4Cl$ (ammonium chloride) and $ZnCl_2$ (zinc chloride). The $MnO_2$ (manganese dioxide) acts as the depolarizer, while the zinc container acts as the anode and a graphite rod acts as the cathode.
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232
ChemistryMediumMCQMHT CET · 2026
What happens during the discharge of a lead storage battery?
A
$SO_2$ is evolved
B
$Pb$ is formed
C
$H_2SO_4$ is consumed
D
$PbSO_4$ is consumed
Solution
(C) During the discharge of a lead storage battery, the following chemical reactions occur:
At anode: $Pb(s) + SO_4^{2-}(aq) \rightarrow PbSO_4(s) + 2e^-$
At cathode: $PbO_2(s) + SO_4^{2-}(aq) + 4H^+(aq) + 2e^- \rightarrow PbSO_4(s) + 2H_2O(l)$
Overall reaction: $Pb(s) + PbO_2(s) + 2H_2SO_4(aq) \rightarrow 2PbSO_4(s) + 2H_2O(l)$
From the overall reaction, it is clear that $H_2SO_4$ is consumed as the battery discharges.
At anode: $Pb(s) + SO_4^{2-}(aq) \rightarrow PbSO_4(s) + 2e^-$
At cathode: $PbO_2(s) + SO_4^{2-}(aq) + 4H^+(aq) + 2e^- \rightarrow PbSO_4(s) + 2H_2O(l)$
Overall reaction: $Pb(s) + PbO_2(s) + 2H_2SO_4(aq) \rightarrow 2PbSO_4(s) + 2H_2O(l)$
From the overall reaction, it is clear that $H_2SO_4$ is consumed as the battery discharges.
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233
ChemistryEasyMCQMHT CET · 2026
Which of the following hot aqueous solutions is used to dip carbon rods of $H_2 - O_2$ fuel cell?
A
$KCl$
B
$KOH$
C
$H_2SO_4$
D
$NH_4Cl$
Solution
(B) In a $H_2 - O_2$ fuel cell, the electrodes are made of porous carbon rods impregnated with a catalyst (like $Pt$ or $Pd$).
These electrodes are dipped into a hot aqueous solution of $KOH$ or $NaOH$, which acts as the electrolyte.
Therefore, the correct option is $KOH$.
These electrodes are dipped into a hot aqueous solution of $KOH$ or $NaOH$, which acts as the electrolyte.
Therefore, the correct option is $KOH$.
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234
ChemistryMediumMCQMHT CET · 2026
What is the change in oxidation number of $Pb$ at the positive electrode of a lead accumulator acting as a galvanic cell?
A
increases by $1$
B
decreases by $1$
C
increases by $2$
D
decreases by $2$
Solution
(D) In a lead accumulator acting as a galvanic cell, the positive electrode is the cathode.
The reduction reaction at the cathode is: $PbO_2 + 4H^+ + SO_4^{2-} + 2e^- \rightarrow PbSO_4 + 2H_2O$.
In $PbO_2$, the oxidation state of $Pb$ is $+4$.
In $PbSO_4$, the oxidation state of $Pb$ is $+2$.
Change in oxidation number = $+2 - (+4) = -2$.
Thus, the oxidation number decreases by $2$.
The reduction reaction at the cathode is: $PbO_2 + 4H^+ + SO_4^{2-} + 2e^- \rightarrow PbSO_4 + 2H_2O$.
In $PbO_2$, the oxidation state of $Pb$ is $+4$.
In $PbSO_4$, the oxidation state of $Pb$ is $+2$.
Change in oxidation number = $+2 - (+4) = -2$.
Thus, the oxidation number decreases by $2$.
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235
ChemistryMediumMCQMHT CET · 2026
Which of the following reactions occurs at the cathode during the recharging of a lead storage battery?
A
$PbSO_4(s) + 2H_2O(\ell) \rightarrow PbO_2(s) + SO_4^{2-}(aq.) + 4H^+(aq.) + 2e^-$
B
$PbO_2(s) + 4H^+(aq.) + SO_4^{2-}(aq.) + 2e^- \rightarrow PbSO_4(s) + 2H_2O(\ell)$
C
$Pb(s) + SO_4^{2-}(aq.) \rightarrow PbSO_4(s) + 2e^-$
D
$PbSO_4(s) + 2e^- \rightarrow Pb(s) + SO_4^{2-}(aq.)$
Solution
(D) During the recharging of a lead storage battery, the cell acts as an electrolytic cell.
At the cathode (negative electrode during recharging), the reduction reaction is: $PbSO_4(s) + 2e^- \rightarrow Pb(s) + SO_4^{2-}(aq.)$.
At the anode (positive electrode during recharging), the oxidation reaction is: $PbSO_4(s) + 2H_2O(\ell) \rightarrow PbO_2(s) + SO_4^{2-}(aq.) + 4H^+(aq.) + 2e^-$.
Therefore, the reaction occurring at the cathode is $PbSO_4(s) + 2e^- \rightarrow Pb(s) + SO_4^{2-}(aq.)$.
At the cathode (negative electrode during recharging), the reduction reaction is: $PbSO_4(s) + 2e^- \rightarrow Pb(s) + SO_4^{2-}(aq.)$.
At the anode (positive electrode during recharging), the oxidation reaction is: $PbSO_4(s) + 2H_2O(\ell) \rightarrow PbO_2(s) + SO_4^{2-}(aq.) + 4H^+(aq.) + 2e^-$.
Therefore, the reaction occurring at the cathode is $PbSO_4(s) + 2e^- \rightarrow Pb(s) + SO_4^{2-}(aq.)$.
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236
ChemistryMediumMCQMHT CET · 2026
Identify the function of a salt bridge in an electrochemical cell.
A
To act as a cathode
B
To act as an anode
C
To connect two half-cells and maintain electrical neutrality
D
To measure the electrode potential
Solution
(C) $1$. $A$ salt bridge is a $U$-shaped tube containing an electrolyte (like $KCl$ or $KNO_3$) in a gel.
$2$. Its primary functions are to connect the two half-cells internally and to maintain electrical neutrality in the solutions of both half-cells by allowing the migration of ions.
$3$. It prevents the accumulation of charge, which would otherwise stop the flow of electrons and the reaction.
$4$. Therefore, option $C$ is the correct function.
$2$. Its primary functions are to connect the two half-cells internally and to maintain electrical neutrality in the solutions of both half-cells by allowing the migration of ions.
$3$. It prevents the accumulation of charge, which would otherwise stop the flow of electrons and the reaction.
$4$. Therefore, option $C$ is the correct function.
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237
ChemistryMediumMCQMHT CET · 2026
Which of the following statements is correct regarding the electrolysis of molten $NaCl$?
A
Pale green $Cl_2$ gas is released at the anode
B
Molten silvery-white sodium metal is deposited at the cathode
C
Decomposition of $NaCl$ into $Na$ metal and $Cl_2$ gas
D
Both options $A$ and $B$
Solution
(D) $1$. During the electrolysis of molten $NaCl$, the dissociation reaction is: $2NaCl(l) \rightarrow 2Na^+(l) + 2Cl^-(l)$.
$2$. At the anode (oxidation): $2Cl^-(l) \rightarrow Cl_2(g) + 2e^-$. Pale green $Cl_2$ gas is evolved.
$3$. At the cathode (reduction): $2Na^+(l) + 2e^- \rightarrow 2Na(l)$. Molten silvery-white sodium metal is deposited.
$4$. Since both statements $A$ and $B$ are correct, the correct option is $D$.
$2$. At the anode (oxidation): $2Cl^-(l) \rightarrow Cl_2(g) + 2e^-$. Pale green $Cl_2$ gas is evolved.
$3$. At the cathode (reduction): $2Na^+(l) + 2e^- \rightarrow 2Na(l)$. Molten silvery-white sodium metal is deposited.
$4$. Since both statements $A$ and $B$ are correct, the correct option is $D$.
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238
ChemistryDifficultMCQMHT CET · 2026
Match List $I$ with List $II$.
Choose the correct answer from the options given below:
| List $I$ (Conversion) | List $II$ (Number of Faraday required) |
| $A$. $1 \text{ mol of } H_2O \text{ to } O_2$ | $I$. $3F$ |
| $B$. $1 \text{ mol of } MnO_4^- \text{ to } Mn^{2+}$ | $II$. $2F$ |
| $C$. $1.5 \text{ mol of } Ca \text{ from molten } CaCl_2$ | $III$. $1F$ |
| $D$. $1 \text{ mol of } FeO \text{ to } Fe_2O_3$ | $IV$. $5F$ |
A
$A-II, B-IV, C-I, D-III$
B
$A-III, B-IV, C-I, D-II$
C
$A-II, B-III, C-I, D-IV$
D
$A-III, B-IV, C-II, D-I$
Solution
(A) . For $H_2O \rightarrow H_2 + \frac{1}{2}O_2$, the reaction is $H_2O \rightarrow \frac{1}{2}O_2 + 2H^+ + 2e^-$. Thus, $2F$ are required.
$B$. For $MnO_4^- \rightarrow Mn^{2+}$, the change in oxidation state of $Mn$ is from $+7$ to $+2$. The reaction is $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$. Thus, $5F$ are required.
$C$. For $Ca^{2+} + 2e^- \rightarrow Ca$, $1 \text{ mol}$ of $Ca$ requires $2F$. So, $1.5 \text{ mol}$ of $Ca$ requires $1.5 \times 2 = 3F$.
$D$. For $FeO \rightarrow \frac{1}{2}Fe_2O_3$, the reaction is $Fe^{2+} \rightarrow Fe^{3+} + e^-$. Thus, $1F$ is required.
Matching: $A-II, B-IV, C-I, D-III$.
$B$. For $MnO_4^- \rightarrow Mn^{2+}$, the change in oxidation state of $Mn$ is from $+7$ to $+2$. The reaction is $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$. Thus, $5F$ are required.
$C$. For $Ca^{2+} + 2e^- \rightarrow Ca$, $1 \text{ mol}$ of $Ca$ requires $2F$. So, $1.5 \text{ mol}$ of $Ca$ requires $1.5 \times 2 = 3F$.
$D$. For $FeO \rightarrow \frac{1}{2}Fe_2O_3$, the reaction is $Fe^{2+} \rightarrow Fe^{3+} + e^-$. Thus, $1F$ is required.
Matching: $A-II, B-IV, C-I, D-III$.
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239
ChemistryDifficultMCQMHT CET · 2026
The standard electrode potential of $Cu^{2+}/Cu$ is $0.34 \text{ V}$ at $298 \text{ K}$. Calculate its electrode potential at the same temperature when the $Cu^{2+}$ ion concentration is $0.1 \text{ M}$. (in $\text{ V}$)
A
$0.64$
B
$0.34$
C
$0.31$
D
$0.37$
Solution
(C) The Nernst equation for the electrode reaction $Cu^{2+} + 2e^- \rightarrow Cu(s)$ is:
$E_{Cu^{2+}/Cu} = E^0_{Cu^{2+}/Cu} - \frac{0.0591}{n} \log \frac{1}{[Cu^{2+}]}$
Here, $n = 2$ and $[Cu^{2+}] = 0.1 \text{ M} = 10^{-1} \text{ M}$.
Substituting the values:
$E = 0.34 - \frac{0.0591}{2} \log \frac{1}{10^{-1}}$
$E = 0.34 - 0.02955 \log(10)$
Since $\log(10) = 1$, we get:
$E = 0.34 - 0.02955 = 0.31045 \text{ V} \approx 0.31 \text{ V}$.
$E_{Cu^{2+}/Cu} = E^0_{Cu^{2+}/Cu} - \frac{0.0591}{n} \log \frac{1}{[Cu^{2+}]}$
Here, $n = 2$ and $[Cu^{2+}] = 0.1 \text{ M} = 10^{-1} \text{ M}$.
Substituting the values:
$E = 0.34 - \frac{0.0591}{2} \log \frac{1}{10^{-1}}$
$E = 0.34 - 0.02955 \log(10)$
Since $\log(10) = 1$, we get:
$E = 0.34 - 0.02955 = 0.31045 \text{ V} \approx 0.31 \text{ V}$.
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240
ChemistryDifficultMCQMHT CET · 2026
Identify which of the following cell reactions is spontaneous under standard state conditions.
A
$Ca(s) + Cd^{2+}(aq) \rightarrow Ca^{2+}(aq) + Cd(s)$ $[E^0_{Ca^{2+}/Ca} = -2.866 \text{ V}, E^0_{Cd^{2+}/Cd} = -0.403 \text{ V}]$
B
$2Br^-(aq) + Sn^{2+}(aq) \rightarrow Br_2(l) + Sn(s)$ $[E^0_{Br_2/Br^-} = 1.08 \text{ V}, E^0_{Sn^{2+}/Sn} = -0.136 \text{ V}]$
C
$2Ag(s) + Ni^{2+}(aq) \rightarrow 2Ag^+(aq) + Ni(s)$ $[E^0_{Ag^+/Ag} = 0.799 \text{ V}, E^0_{Ni^{2+}/Ni} = -0.257 \text{ V}]$
D
$2Au(s) + Zn^{2+}(aq) \rightarrow 2Au^+(aq) + Zn(s)$ $[E^0_{Au^+/Au} = 1.68 \text{ V}, E^0_{Zn^{2+}/Zn} = -0.763 \text{ V}]$
Solution
(A) reaction is spontaneous if $E^0_{cell} > 0$. $E^0_{cell} = E^0_{cathode} - E^0_{anode}$.
For $(A)$: $E^0_{cell} = (-0.403) - (-2.866) = +2.463 \text{ V}$. Since $E^0_{cell} > 0$, it is spontaneous.
For $(B)$: $E^0_{cell} = (-0.136) - (1.08) = -1.216 \text{ V}$. Non-spontaneous.
For $(C)$: $E^0_{cell} = (-0.257) - (0.799) = -1.056 \text{ V}$. Non-spontaneous.
For $(D)$: $E^0_{cell} = (-0.763) - (1.68) = -2.443 \text{ V}$. Non-spontaneous.
For $(A)$: $E^0_{cell} = (-0.403) - (-2.866) = +2.463 \text{ V}$. Since $E^0_{cell} > 0$, it is spontaneous.
For $(B)$: $E^0_{cell} = (-0.136) - (1.08) = -1.216 \text{ V}$. Non-spontaneous.
For $(C)$: $E^0_{cell} = (-0.257) - (0.799) = -1.056 \text{ V}$. Non-spontaneous.
For $(D)$: $E^0_{cell} = (-0.763) - (1.68) = -2.443 \text{ V}$. Non-spontaneous.
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241
ChemistryEasyMCQMHT CET · 2026
Which of the following statements is correct regarding the cathode of an electrochemical cell?
A
Oxidation occurs at the cathode.
B
Reduction occurs at the cathode.
C
It is usually denoted by a negative sign.
D
It is usually made up of non-conducting material.
Solution
(B) In an electrochemical cell, the cathode is the electrode where reduction (gain of electrons) takes place.
In a galvanic cell, the cathode is positive, while in an electrolytic cell, the cathode is negative.
Therefore, the statement 'Reduction occurs at the cathode' is always correct.
In a galvanic cell, the cathode is positive, while in an electrolytic cell, the cathode is negative.
Therefore, the statement 'Reduction occurs at the cathode' is always correct.
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242
ChemistryMediumMCQMHT CET · 2026
Which of the following reactions is possible at the anode?
A
$F_2 + 2e^- \rightarrow 2F^-$
B
$2H^+ + \frac{1}{2} O_2 + 2e^- \rightarrow H_2O$
C
$Fe^{2+} \rightarrow Fe^{3+} + 1e^-$
D
$Cu^{2+} + 2e^- \rightarrow Cu(s)$
Solution
(C) $1$. The anode is the electrode where oxidation occurs.
$2$. Oxidation is defined as the loss of electrons.
$3$. In option $(A)$, $F_2$ gains electrons (reduction).
$4$. In option $(B)$, $H^+$ and $O_2$ gain electrons (reduction).
$5$. In option $(C)$, $Fe^{2+}$ loses an electron to form $Fe^{3+}$, which is an oxidation process.
$6$. In option $(D)$, $Cu^{2+}$ gains electrons (reduction).
$7$. Therefore, the reaction in option $(C)$ is the only one that represents oxidation and can occur at the anode.
$2$. Oxidation is defined as the loss of electrons.
$3$. In option $(A)$, $F_2$ gains electrons (reduction).
$4$. In option $(B)$, $H^+$ and $O_2$ gain electrons (reduction).
$5$. In option $(C)$, $Fe^{2+}$ loses an electron to form $Fe^{3+}$, which is an oxidation process.
$6$. In option $(D)$, $Cu^{2+}$ gains electrons (reduction).
$7$. Therefore, the reaction in option $(C)$ is the only one that represents oxidation and can occur at the anode.
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243
ChemistryDifficultMCQMHT CET · 2026
The $E^0_{cell}$ of the cell $Al(s)|Al^{3+}(1M)||Pb^{2+}(1M)|Pb(s)$ is $1.5 \text{ V}$. If $E^0_{Pb^{2+}/Pb}$ is $-0.14 \text{ V}$, then $E^0_{Al^{3+}/Al}$ will be: (in $\text{ V}$)
A
$1.64$
B
$-1.64$
C
$1.36$
D
$-1.36$
Solution
(B) The standard cell potential is given by the formula: $E^0_{cell} = E^0_{cathode} - E^0_{anode}$.
Here, the cathode is $Pb^{2+}/Pb$ and the anode is $Al^{3+}/Al$.
Given: $E^0_{cell} = 1.5 \text{ V}$ and $E^0_{cathode} = -0.14 \text{ V}$.
Substituting the values: $1.5 \text{ V} = -0.14 \text{ V} - E^0_{anode}$.
Rearranging the equation: $E^0_{anode} = -0.14 \text{ V} - 1.5 \text{ V}$.
Therefore, $E^0_{Al^{3+}/Al} = -1.64 \text{ V}$.
Here, the cathode is $Pb^{2+}/Pb$ and the anode is $Al^{3+}/Al$.
Given: $E^0_{cell} = 1.5 \text{ V}$ and $E^0_{cathode} = -0.14 \text{ V}$.
Substituting the values: $1.5 \text{ V} = -0.14 \text{ V} - E^0_{anode}$.
Rearranging the equation: $E^0_{anode} = -0.14 \text{ V} - 1.5 \text{ V}$.
Therefore, $E^0_{Al^{3+}/Al} = -1.64 \text{ V}$.
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244
ChemistryDifficultMCQMHT CET · 2026
Cells $A$ and $B$ contain aqueous solutions of ferrous chloride $(FeCl_2)$ and ferric chloride $(FeCl_3)$ respectively. If the same quantity of electricity is passed through both cells, what is the ratio of the mass of iron deposited in cell $A$ to that in cell $B$?
A
$1:1$
B
$2:1$
C
$3:1$
D
$3:2$
Solution
(D) According to Faraday's second law of electrolysis, the mass $(m)$ of a substance deposited is given by $m = \frac{M \cdot Q}{n \cdot F}$, where $M$ is the molar mass, $Q$ is the quantity of electricity, $n$ is the valency factor, and $F$ is Faraday's constant.
Since $Q$ and $F$ are constant, $m \propto \frac{M}{n}$.
For cell $A$ $(FeCl_2)$, the iron is in the $Fe^{2+}$ state, so $n_A = 2$.
For cell $B$ $(FeCl_3)$, the iron is in the $Fe^{3+}$ state, so $n_B = 3$.
The ratio of the mass of iron deposited is $\frac{m_A}{m_B} = \frac{M/n_A}{M/n_B} = \frac{n_B}{n_A} = \frac{3}{2} = 3:2$.
Since $Q$ and $F$ are constant, $m \propto \frac{M}{n}$.
For cell $A$ $(FeCl_2)$, the iron is in the $Fe^{2+}$ state, so $n_A = 2$.
For cell $B$ $(FeCl_3)$, the iron is in the $Fe^{3+}$ state, so $n_B = 3$.
The ratio of the mass of iron deposited is $\frac{m_A}{m_B} = \frac{M/n_A}{M/n_B} = \frac{n_B}{n_A} = \frac{3}{2} = 3:2$.
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245
ChemistryEasyMCQMHT CET · 2026
Identify the reaction from the following that corresponds to the standard electrode potential of the $Cu^{+2}/Cu$ electrode being $0.34 \text{ V}$ with respect to the $SHE$?
A
$Cu \rightarrow Cu^{+2} + 2e^-$
B
$Cu^{+2} + 2e^- \rightarrow Cu$
C
$Cu^+ \rightarrow Cu^{+2} + e^-$
D
$Cu^{+3} \rightarrow Cu^{+2} + e^-$
Solution
(B) The standard electrode potential $E^\circ$ is defined for the reduction half-reaction.
For the $Cu^{+2}/Cu$ electrode, the reduction reaction is the gain of electrons by the copper ion to form metallic copper.
The reaction is: $Cu^{+2}(aq) + 2e^- \rightarrow Cu(s)$.
Thus, option $B$ represents the correct reduction reaction.
For the $Cu^{+2}/Cu$ electrode, the reduction reaction is the gain of electrons by the copper ion to form metallic copper.
The reaction is: $Cu^{+2}(aq) + 2e^- \rightarrow Cu(s)$.
Thus, option $B$ represents the correct reduction reaction.
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246
ChemistryMediumMCQMHT CET · 2026
Which of the following reactions takes place at the anode during the electrolysis of molten $NaCl$?
A
$Na^+(l) + e^- \rightarrow Na(l)$
B
$Na(s) \rightarrow Na^+(l) + e^-$
C
$Cl_2(g) + 2e^- \rightarrow 2Cl^-(l)$
D
$2Cl^-(l) \rightarrow Cl_2(g) + 2e^-$
Solution
(D) $1$. During the electrolysis of molten $NaCl$, the electrolyte dissociates into $Na^+$ and $Cl^-$ ions.
$2$. The anode is the positively charged electrode, which attracts negatively charged ions (anions).
$3$. The chloride ions $(Cl^-)$ migrate to the anode.
$4$. At the anode, oxidation occurs, where $Cl^-$ ions lose electrons to form chlorine gas $(Cl_2)$: $2Cl^-(l) \rightarrow Cl_2(g) + 2e^-$.
$5$. Therefore, option $D$ is the correct reaction.
$2$. The anode is the positively charged electrode, which attracts negatively charged ions (anions).
$3$. The chloride ions $(Cl^-)$ migrate to the anode.
$4$. At the anode, oxidation occurs, where $Cl^-$ ions lose electrons to form chlorine gas $(Cl_2)$: $2Cl^-(l) \rightarrow Cl_2(g) + 2e^-$.
$5$. Therefore, option $D$ is the correct reaction.
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247
ChemistryDifficultMCQMHT CET · 2026
What mass of silver (Atomic mass = $108 \text{ g/mol}$) is deposited by a quantity of electricity that displaces $5600 \text{ mL}$ of $O_2$ gas at $STP$ (in $\text{ g}$)?
A
$5.4$
B
$10.8$
C
$54.0$
D
$108.0$
Solution
(D) Step $1$: Calculate moles of $O_2$ gas at $STP$. Moles = $5600 \text{ mL} / 22400 \text{ mL/mol} = 0.25 \text{ mol}$.
Step $2$: Calculate equivalents of $O_2$. The reaction is $2H_2O \rightarrow O_2 + 4H^+ + 4e^-$. The $n$-factor for $O_2$ is $4$. Equivalents = $0.25 \times 4 = 1$.
Step $3$: According to Faraday's law, equivalents of $Ag$ deposited = equivalents of $O_2$ displaced = $1$.
Step $4$: Mass of $Ag$ = Equivalents $\times$ Equivalent mass = $1 \times 108 \text{ g/mol} = 108 \text{ g}$.
Step $2$: Calculate equivalents of $O_2$. The reaction is $2H_2O \rightarrow O_2 + 4H^+ + 4e^-$. The $n$-factor for $O_2$ is $4$. Equivalents = $0.25 \times 4 = 1$.
Step $3$: According to Faraday's law, equivalents of $Ag$ deposited = equivalents of $O_2$ displaced = $1$.
Step $4$: Mass of $Ag$ = Equivalents $\times$ Equivalent mass = $1 \times 108 \text{ g/mol} = 108 \text{ g}$.
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248
ChemistryDifficultMCQMHT CET · 2026
If the $emf$ of the cell $Cu(s) | Cu^{2+}(1 \text{ M}) || Ag^+(1 \text{ M}) | Ag(s)$ is $0.463 \text{ V}$ at $25^{\circ} \text{C}$ and the standard electrode potential of the $Cu$ electrode is $0.337 \text{ V}$, find the standard electrode potential of the $Ag$ electrode. (in $\text{ V}$)
A
$0.128$
B
$-0.128$
C
$0.8$
D
$-0.8$
Solution
(C) The standard cell potential $E^0_{cell}$ is given by the difference between the standard reduction potentials of the cathode and the anode:
$E^0_{cell} = E^0_{cathode} - E^0_{anode}$
Here, the cathode is $Ag$ and the anode is $Cu$, so:
$E^0_{cell} = E^0_{Ag^+/Ag} - E^0_{Cu^{2+}/Cu}$
Given $E^0_{cell} = 0.463 \text{ V}$ and $E^0_{Cu^{2+}/Cu} = 0.337 \text{ V}$:
$0.463 \text{ V} = E^0_{Ag^+/Ag} - 0.337 \text{ V}$
$E^0_{Ag^+/Ag} = 0.463 \text{ V} + 0.337 \text{ V} = 0.800 \text{ V}$
Thus, the standard potential of the $Ag$ electrode is $0.8 \text{ V}$.
$E^0_{cell} = E^0_{cathode} - E^0_{anode}$
Here, the cathode is $Ag$ and the anode is $Cu$, so:
$E^0_{cell} = E^0_{Ag^+/Ag} - E^0_{Cu^{2+}/Cu}$
Given $E^0_{cell} = 0.463 \text{ V}$ and $E^0_{Cu^{2+}/Cu} = 0.337 \text{ V}$:
$0.463 \text{ V} = E^0_{Ag^+/Ag} - 0.337 \text{ V}$
$E^0_{Ag^+/Ag} = 0.463 \text{ V} + 0.337 \text{ V} = 0.800 \text{ V}$
Thus, the standard potential of the $Ag$ electrode is $0.8 \text{ V}$.
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249
ChemistryDifficultMCQMHT CET · 2026
If standard reduction potentials $(E^0)$ of $(Al^{+3}(aq)|Al(s))$, $(Fe^{+2}(aq)|Fe(s))$, $(Cu^{+2}(aq)|Cu(s))$ and $(Ag^{+1}(aq)|Ag(s))$ are $-1.66 \text{ V}$, $-0.44 \text{ V}$, $+0.34 \text{ V}$ and $+0.79 \text{ V}$ respectively, which of the following reactions is non-spontaneous?
A
$2Ag(s) + Fe^{+2}(aq) \rightarrow 2Ag^{+}(aq) + Fe(s)$
B
$2Al(s) + 3Cu^{+2}(aq) \rightarrow 2Al^{+3}(aq) + 3Cu(s)$
C
$Fe(s) + Cu^{+2}(aq) \rightarrow Fe^{+2}(aq) + Cu(s)$
D
$2Al(s) + 3Fe^{+2}(aq) \rightarrow 2Al^{+3}(aq) + 3Fe(s)$
Solution
(A) reaction is non-spontaneous if the standard cell potential $E^0_{cell} < 0$.
For option $A$: $E^0_{cell} = E^0_{cathode} - E^0_{anode} = E^0_{Fe^{+2}/Fe} - E^0_{Ag^{+}/Ag} = -0.44 - 0.79 = -1.23 \text{ V}$. Since $E^0_{cell} < 0$, this reaction is non-spontaneous.
For option $B$: $E^0_{cell} = 0.34 - (-1.66) = 2.00 \text{ V} > 0$ (Spontaneous).
For option $C$: $E^0_{cell} = 0.34 - (-0.44) = 0.78 \text{ V} > 0$ (Spontaneous).
For option $D$: $E^0_{cell} = -0.44 - (-1.66) = 1.22 \text{ V} > 0$ (Spontaneous).
For option $A$: $E^0_{cell} = E^0_{cathode} - E^0_{anode} = E^0_{Fe^{+2}/Fe} - E^0_{Ag^{+}/Ag} = -0.44 - 0.79 = -1.23 \text{ V}$. Since $E^0_{cell} < 0$, this reaction is non-spontaneous.
For option $B$: $E^0_{cell} = 0.34 - (-1.66) = 2.00 \text{ V} > 0$ (Spontaneous).
For option $C$: $E^0_{cell} = 0.34 - (-0.44) = 0.78 \text{ V} > 0$ (Spontaneous).
For option $D$: $E^0_{cell} = -0.44 - (-1.66) = 1.22 \text{ V} > 0$ (Spontaneous).
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250
ChemistryMediumMCQMHT CET · 2026
Which of the following net cell reactions occurs in a galvanic cell containing a cadmium electrode and a standard hydrogen electrode? Given: $E^0(Cd^{2+}(aq)|Cd(s)) = -0.403 \text{ V}$.
A
$H_2(g) + Cd^{2+}(aq) \rightarrow 2H^+(aq) + Cd(s)$
B
$Cd(s) + 2H^+(aq) \rightarrow Cd^{2+}(aq) + H_2(g)$
C
$2H_2(g) + Cd^{2+}(aq) \rightarrow 4H^+(aq) + Cd(s)$
D
$2Cd(s) + 2H^+(aq) \rightarrow 2Cd^{2+}(aq) + H_2(g)$
Solution
(B) Step $1$: Identify the standard electrode potentials. For the standard hydrogen electrode $(SHE)$, $E^0(H^+|H_2) = 0.00 \text{ V}$. For the cadmium electrode, $E^0(Cd^{2+}|Cd) = -0.403 \text{ V}$.
Step $2$: Determine the anode and cathode. Since $E^0(Cd^{2+}|Cd) < E^0(H^+|H_2)$, the cadmium electrode acts as the anode (oxidation) and the hydrogen electrode acts as the cathode (reduction).
Step $3$: Write the half-reactions. Anode: $Cd(s) \rightarrow Cd^{2+}(aq) + 2e^-$. Cathode: $2H^+(aq) + 2e^- \rightarrow H_2(g)$.
Step $4$: Combine the half-reactions to get the net cell reaction: $Cd(s) + 2H^+(aq) \rightarrow Cd^{2+}(aq) + H_2(g)$.
Step $2$: Determine the anode and cathode. Since $E^0(Cd^{2+}|Cd) < E^0(H^+|H_2)$, the cadmium electrode acts as the anode (oxidation) and the hydrogen electrode acts as the cathode (reduction).
Step $3$: Write the half-reactions. Anode: $Cd(s) \rightarrow Cd^{2+}(aq) + 2e^-$. Cathode: $2H^+(aq) + 2e^- \rightarrow H_2(g)$.
Step $4$: Combine the half-reactions to get the net cell reaction: $Cd(s) + 2H^+(aq) \rightarrow Cd^{2+}(aq) + H_2(g)$.
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