MHT CET 2026 Chemistry Question Paper with Answer and Solution

(B) The compressibility factor $Z$ is defined as $Z = \frac{PV}{nRT}$.
Given values are: $P = 40 \text{ atm}$, $V = 0.4 \text{ dm}^3$, $n = 1 \text{ mol}$, $T = 300 \text{ K}$, and $R = 0.082 \text{ dm}^3 \text{ atm K}^{-1} \text{ mol}^{-1}$.
Substituting these values into the formula:
$Z = \frac{40 \times 0.4}{1 \times 0.082 \times 300}$
$Z = \frac{16}{24.6}$
$Z \approx 0.65$.

(C) The average kinetic energy of gas molecules is given by the formula $KE_{avg} = \frac{3}{2} k_B T$, where $k_B$ is the Boltzmann constant and $T$ is the absolute temperature.
Since $k_B$ is a constant, the average kinetic energy is directly proportional to the absolute temperature $T$.

(C) $1$. An ideal gas is a theoretical model that obeys gas laws under all conditions of temperature and pressure.
$2$. The kinetic molecular theory assumes that ideal gas molecules have no volume and no intermolecular forces of attraction.
$3$. Since these conditions cannot be met by real gases, an ideal gas does not exist in reality.

(B) The partial pressure of a gas is given by $P_i = x_i P_{total}$, where $x_i$ is the mole fraction.
For the partial pressures to be equal $(P_{Ne} = P_{CH_4})$, their mole fractions must be equal $(x_{Ne} = x_{CH_4})$.
Since $x_i = \frac{n_i}{n_{total}}$, equal mole fractions imply equal number of moles $(n_{Ne} = n_{CH_4})$.
Let the number of moles be $n$. The mass of $Ne$ is $m_{Ne} = n \times M_{Ne} = n \times 20 \ \text{g/mol}$.
The mass of $CH_4$ is $m_{CH_4} = n \times M_{CH_4} = n \times 16 \ \text{g/mol}$.
The ratio by mass is $\frac{m_{Ne}}{m_{CH_4}} = \frac{20n}{16n} = \frac{5}{4}$ or $5:4$.

(A) Let the total mass of the mixture be $100 \text{ g}$.
Mass of $He$ $(m_{He})$ = $20 \text{ g}$, Mass of $Ne$ $(m_{Ne})$ = $80 \text{ g}$.
Molar mass of $He$ $(M_{He})$ = $4 \text{ g/mol}$, Molar mass of $Ne$ $(M_{Ne})$ = $20 \text{ g/mol}$.
Number of moles of $He$ $(n_{He})$ = $\frac{20 \text{ g}}{4 \text{ g/mol}} = 5 \text{ mol}$.
Number of moles of $Ne$ $(n_{Ne})$ = $\frac{80 \text{ g}}{20 \text{ g/mol}} = 4 \text{ mol}$.
According to Dalton's Law, the ratio of partial pressures is equal to the ratio of the number of moles: $\frac{P_{He}}{P_{Ne}} = \frac{n_{He}}{n_{Ne}} = \frac{5}{4}$.

(B) Step $1$: Calculate the total number of moles in the mixture.
$n_{\text{total}} = n_{N_2} + n_{O_2} + n_{H_2} = 0.5 + 1.0 + 1.5 = 3.0 \text{ mol}$.
Step $2$: Calculate the mole fraction of $O_2$ $(x_{O_2})$.
$x_{O_2} = \frac{n_{O_2}}{n_{\text{total}}} = \frac{1.0}{3.0} = \frac{1}{3}$.
Step $3$: Calculate the partial pressure of $O_2$ using Dalton's Law.
$P_{O_2} = x_{O_2} \times P_{\text{total}} = \frac{1}{3} \times 18 \text{ bar} = 6.0 \text{ bar}$.

(C) According to Boyle's Law, for a fixed amount of gas at constant temperature, $P_1V_1 = P_2V_2$.
Given: $P_1 = 1 \text{ atm}$, $V_1 = 10.0 \text{ cm}^3$, $P_2 = 3.5 \text{ atm}$.
Substituting the values: $1 \text{ atm} \times 10.0 \text{ cm}^3 = 3.5 \text{ atm} \times V_2$.
$V_2 = \frac{10.0 \text{ cm}^3 \times 1 \text{ atm}}{3.5 \text{ atm}}$.
$V_2 \approx 2.86 \text{ cm}^3$.

(A) Using the ideal gas equation: $PV = nRT$
Given: $P = 24.6 \text{ atm}$, $V = 5.0 \text{ L}$, $n = 2 \text{ mol}$, $R = 0.0821 \text{ L atm K}^{-1} \text{ mol}^{-1}$
Substitute the values: $24.6 \times 5.0 = 2 \times 0.0821 \times T$
$123 = 0.1642 \times T$
$T = 123 / 0.1642 \approx 749.08 \text{ K}$
Convert to Celsius: $T(^\circ \text{C}) = T(\text{K}) - 273.15$
$T \approx 749.08 - 273.15 = 475.93^\circ \text{C} \approx 476^\circ \text{C}$

(B) Given: $P_1 = 2 \ atm$, $V_1 = 0.35 \ dm^3$, $T_1 = 290 \ K$.
At $STP$: $P_2 = 1 \ atm$, $T_2 = 273.15 \ K$ (commonly taken as $273 \ K$).
Using the combined gas law: $\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$.
Substituting the values: $\frac{2 \times 0.35}{290} = \frac{1 \times V_2}{273}$.
$V_2 = \frac{0.7 \times 273}{290} \approx 0.659 \ dm^3$.
Rounding to two decimal places, $V_2 \approx 0.66 \ dm^3$.

(A) Using the combined gas law: $\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$
Given: $P_1 = 1 \text{ atm}$, $V_1 = 5 \text{ L}$, $T_1 = 300 \text{ K}$, $V_2 = 2 \text{ L}$, $T_2 = 500 \text{ K}$.
Substitute the values: $\frac{1 \times 5}{300} = \frac{P_2 \times 2}{500}$
$P_2 = \frac{1 \times 5 \times 500}{300 \times 2} = \frac{2500}{600} = \frac{25}{6} \approx 4.17 \text{ atm}$.

(B) According to Boyle's Law, at constant temperature, $P_1V_1 = P_2V_2$.
Given: $P_1 = 500 \text{ mm Hg}$, $V_1 = 200 \text{ mL}$, $P_2 = 800 \text{ mm Hg}$.
Substituting the values: $500 \times 200 = 800 \times V_2$.
$V_2 = \frac{500 \times 200}{800} \text{ mL}$.
$V_2 = \frac{100000}{800} \text{ mL} = 125 \text{ mL}$.

(C) According to Charles' Law, for a fixed mass of gas at constant pressure, the volume is directly proportional to the absolute temperature.
$V_t = V_0 \left(1 + \frac{t}{273.15}\right)$.
Using the approximation $273$ for $273.15$:
$V_t = V_0 \left(1 + \frac{t}{273}\right)$.
$V_t = V_0 \left(\frac{273 + t}{273}\right)$.
Therefore, the ratio $\frac{V_t}{V_0} = \frac{273 + t}{273}$.

(A) The change in entropy of the surroundings is given by the formula: $\Delta S_{\text{surr}} = -\frac{\Delta H_{\text{sys}}}{T}$.
Given: $\Delta H_{\text{sys}} = -286 \text{ kJ mol}^{-1} = -286000 \text{ J mol}^{-1}$ and $T = 298 \text{ K}$.
Substituting the values: $\Delta S_{\text{surr}} = -\frac{-286000 \text{ J mol}^{-1}}{298 \text{ K}}$.
$\Delta S_{\text{surr}} = \frac{286000}{298} \text{ J K}^{-1} \text{ mol}^{-1} \approx 959.73 \text{ J K}^{-1} \text{ mol}^{-1}$.
Thus, the correct option is $A$.

(C) $1$. Molar mass of $CO = 12 + 16 = 28 \text{ g/mol}$.
$2$. Moles of $CO = \frac{\text{mass}}{\text{molar mass}} = \frac{6 \text{ g}}{28 \text{ g/mol}} \approx 0.2143 \text{ mol}$.
$3$. From the balanced equation, $4 \text{ moles}$ of $CO$ release $1200 \text{ kJ}$ of energy.
$4$. Enthalpy change for $0.2143 \text{ moles}$ of $CO = \frac{-1200 \text{ kJ}}{4 \text{ mol}} \times 0.2143 \text{ mol} = -300 \times 0.2143 \text{ kJ} \approx -64.29 \text{ kJ}$.

(B) $1$. The molar mass of $CO_2$ is $12 + 2 \times 16 = 44 \text{ g/mol}$.
$2$. The number of moles of $CO_2$ formed is $n = \frac{\text{mass}}{\text{molar mass}} = \frac{35.2 \text{ g}}{44 \text{ g/mol}} = 0.8 \text{ mol}$.
$3$. The heat released for $1 \text{ mol}$ of $CO_2$ is $393.5 \text{ kJ}$.
$4$. The heat released for $0.8 \text{ mol}$ of $CO_2$ is $0.8 \times 393.5 \text{ kJ} = 314.8 \text{ kJ}$.
$5$. Since the reaction is exothermic, the heat released is $314.8 \text{ kJ}$, which is approximately $-315 \text{ kJ}$ (as per the sign convention for heat released).

(A) The hydrogenation reaction is: $C_6H_{10}(l) + H_2(g) \rightarrow C_6H_{12}(l)$.
Using the enthalpy of combustion data: $\Delta H_{\text{hydrogenation}} = \sum \Delta H_c(\text{reactants}) - \sum \Delta H_c(\text{products})$.
$\Delta H_{\text{hydrogenation}} = [\Delta H_c(C_6H_{10}) + \Delta H_c(H_2)] - [\Delta H_c(C_6H_{12})]$.
$\Delta H_{\text{hydrogenation}} = [(-3800) + (-241)] - (-3920) \text{ kJ mol}^{-1}$.
$\Delta H_{\text{hydrogenation}} = -4041 + 3920 = -121 \text{ kJ mol}^{-1}$.

(A) The standard enthalpy change of a reaction is calculated by subtracting the sum of the standard enthalpies of formation of the reactants from the sum of the standard enthalpies of formation of the products.
Formula: $\Delta_r H^\circ = \sum \Delta_f H^\circ(\text{products}) - \sum \Delta_f H^\circ(\text{reactants})$
Therefore, option $A$ is correct.

(C) $1$. An endothermic process is one that absorbs heat from the surroundings.
$2$. Mixing $NaOH$ and $HCl$ is a neutralization reaction, which is exothermic.
$3$. The $Haber$ process $(N_2 + 3H_2 \rightleftharpoons 2NH_3)$ is exothermic.
$4$. Freezing of water is an exothermic process as it releases latent heat.
$5$. Melting of ice requires the absorption of latent heat from the surroundings, making it an endothermic process.

(C) The combustion reaction of carbon monoxide is: $CO(g) + \frac{1}{2} O_2(g) \rightarrow CO_2(g)$.
The standard enthalpy of combustion $\Delta_c H^{\circ}$ is given by the formula: $\Delta_c H^{\circ} = \sum \Delta_f H^{\circ}(\text{products}) - \sum \Delta_f H^{\circ}(\text{reactants})$.
Substituting the given values: $\Delta_c H^{\circ} = [\Delta_f H^{\circ}(CO_2)] - [\Delta_f H^{\circ}(CO) + \frac{1}{2} \Delta_f H^{\circ}(O_2)]$.
Since $\Delta_f H^{\circ}(O_2) = 0 \text{ kJ mol}^{-1}$ for an element in its standard state, we have: $\Delta_c H^{\circ} = -393 - (-110 + 0) \text{ kJ mol}^{-1}$.
$\Delta_c H^{\circ} = -393 + 110 = -283 \text{ kJ mol}^{-1}$.

(B) The standard enthalpy change of reaction is given by: $\Delta_r H^{\circ} = \sum \Delta_f H^{\circ}(\text{products}) - \sum \Delta_f H^{\circ}(\text{reactants})$.
For the reaction $2C_2H_6(g) + 7O_2(g) \rightarrow 4CO_2(g) + 6H_2O(l)$:
$\Delta_r H^{\circ} = [4 \times \Delta_f H^{\circ}(CO_2) + 6 \times \Delta_f H^{\circ}(H_2O)] - [2 \times \Delta_f H^{\circ}(C_2H_6) + 7 \times \Delta_f H^{\circ}(O_2)]$.
Since $\Delta_f H^{\circ}(O_2) = 0 \text{ kJ mol}^{-1}$ (standard state of an element):
$\Delta_r H^{\circ} = [4(-390) + 6(-285)] - [2(-85) + 7(0)]$.
$\Delta_r H^{\circ} = [-1560 - 1710] - [-170]$.
$\Delta_r H^{\circ} = -3270 + 170 = -3100 \text{ kJ}$.

(B) Step $1$: Hess's Law of Constant Heat Summation states that if a reaction takes place in several steps, then its standard reaction enthalpy is the sum of the standard enthalpies of the intermediate reactions into which the overall reaction may be divided at the same temperature.
Step $2$: The statement provided describes this principle exactly.
Step $3$: Therefore, the correct law is Hess's law.

(A) The formation reaction of methane is: $C(s) + 2H_2(g) \rightarrow CH_4(g)$.
Given combustion reactions:
$(1)$ $C(s) + O_2(g) \rightarrow CO_2(g)$; $\Delta H_1 = -390 \text{ kJ/mol}$
$(2)$ $H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l)$; $\Delta H_2 = -285 \text{ kJ/mol}$
$(3)$ $CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)$; $\Delta H_3 = -890 \text{ kJ/mol}$
To get the formation reaction, calculate: $\Delta H_f = \Delta H_1 + 2(\Delta H_2) - \Delta H_3$.
$\Delta H_f = -390 + 2(-285) - (-890)$
$\Delta H_f = -390 - 570 + 890$
$\Delta H_f = -960 + 890 = -70 \text{ kJ/mol}$.

(B) $1$. The internal energy $\Delta U$ of an ideal gas is a function of temperature only, i.e., $\Delta U = f(T)$.
$2$. For an isothermal process, the temperature remains constant, so $\Delta T = 0$.
$3$. Since $\Delta T = 0$, the change in internal energy $\Delta U$ is equal to $0$.

(A) For a reversible isothermal expansion of an ideal gas, the work done $w$ is given by the formula:
$w = -nRT \ln\left(\frac{P_1}{P_2}\right)$
Given:
$n = 1 \text{ mol}$
$R = 8.314 \text{ J K}^{-1} \text{ mol}^{-1}$
$T = 300 \text{ K}$
$P_1 = 10 \text{ bar}$
$P_2 = 1 \text{ bar}$
Substituting the values:
$w = -(1 \text{ mol}) \times (8.314 \text{ J K}^{-1} \text{ mol}^{-1}) \times (300 \text{ K}) \times \ln\left(\frac{10}{1}\right)$
$w = -2494.2 \times \ln(10)$
Since $\ln(10) \approx 2.303$:
$w = -2494.2 \times 2.303 \approx -5744.14 \text{ J}$
Converting to kilojoules:
$w \approx -5.744 \text{ kJ}$
Thus, the correct option is $A$.

(C) $1$. The internal energy $(U)$ of an ideal gas is a function of temperature $(T)$ only, i.e., $U = f(T)$.
$2$. In an isothermal process, the temperature of the system remains constant $(dT = 0)$.
$3$. Since the temperature does not change, the internal energy of the system remains constant $(\Delta U = 0)$.
$4$. Therefore, the correct option is $C$.

(A) Step $1$: Calculate the work done by the system using $w = -P_{ext} \Delta V$.
Step $2$: Convert units: $1 \text{ bar} = 10^5 \text{ Pa}$ and $1 \text{ dm}^3 = 10^{-3} \text{ m}^3$.
Step $3$: $\Delta V = 1.25 \text{ dm}^3 - 1 \text{ dm}^3 = 0.25 \text{ dm}^3 = 0.25 \times 10^{-3} \text{ m}^3$.
Step $4$: $w = -(10^5 \text{ Pa}) \times (0.25 \times 10^{-3} \text{ m}^3) = -25 \text{ J}$.
Step $5$: Apply the first law of thermodynamics: $\Delta U = q + w$.
Step $6$: Given $q = +150 \text{ J}$ (heat supplied to the system).
Step $7$: $\Delta U = 150 \text{ J} - 25 \text{ J} = 125 \text{ J}$.

(A) $1$. Properties that depend on the amount of matter are called extensive properties (e.g., $Heat$, $Enthalpy$, $Entropy$, $Free \ energy$).
$2$. Properties that depend on the path taken are called path functions.
$3$. $Heat$ $(q)$ and $Work$ $(w)$ are path functions, whereas $Free \ energy$, $Enthalpy$, and $Entropy$ are state functions (they depend only on the initial and final states, not the path).
$4$. Since $Heat$ is both an extensive property (depends on the amount of matter) and a path function, it is the correct answer.

(C) $1$. $A$ state function is a property whose value depends only on the current state of the system and not on the path taken to reach that state.
$2$. $Volume$ $(V)$, $Pressure$ $(P)$, and $Temperature$ $(T)$ are state functions because they depend only on the state variables of the system.
$3$. $Work$ $(w)$ and $Heat$ $(q)$ are path functions, meaning their values depend on the process or path taken to change the state of the system.
$4$. Therefore, $Work$ is not a state function.

(B) The balanced chemical equation is $NH_{3(g)} + HCl_{(g)} \rightarrow NH_4Cl_{(s)}$.
$\Delta n_g$ is defined as the difference between the sum of stoichiometric coefficients of gaseous products and gaseous reactants.
$\Delta n_g = \sum n_{p(g)} - \sum n_{r(g)}$.
Here, the number of moles of gaseous products is $0$ (since $NH_4Cl$ is solid).
The number of moles of gaseous reactants is $1 + 1 = 2$.
Therefore, $\Delta n_g = 0 - 2 = -2$.

(B) The relationship between standard Gibbs free energy change and equilibrium constant is given by: $\Delta G^\circ = -RT \ln K_p$.
Given: $R = 8.314 \text{ J K}^{-1} \text{mol}^{-1}$, $T = 298 \text{ K}$, $K_p = 2 \times 10^{17}$.
$\Delta G^\circ = -8.314 \times 298 \times \ln(2 \times 10^{17})$.
$\Delta G^\circ = -8.314 \times 298 \times (\ln 2 + \ln 10^{17})$.
$\Delta G^\circ = -8.314 \times 298 \times (0.693 + 17 \times 2.303)$.
$\Delta G^\circ = -8.314 \times 298 \times (0.693 + 39.151) = -8.314 \times 298 \times 39.844$.
$\Delta G^\circ \approx -98716 \text{ J mol}^{-1} = -98.716 \text{ kJ mol}^{-1}$.

(A) The dissociation of $BaSO_4$ is: $BaSO_4(s) \rightleftharpoons Ba^{2+}(aq) + SO_4^{2-}(aq)$.
Let solubility be $S \text{ mol/L}$. Then $K_{sp} = [Ba^{2+}][SO_4^{2-}] = S^2$.
$S = \sqrt{K_{sp}} = \sqrt{1.0 \times 10^{-10}} = 1.0 \times 10^{-5} \text{ mol/L}$.
Since $1 \text{ mol/L} = 1 \text{ mol/dm}^3$, $S = 1.0 \times 10^{-5} \text{ mol/dm}^3$.
Solubility in $\text{g/dm}^3 = S \times \text{Molar mass} = 1.0 \times 10^{-5} \text{ mol/dm}^3 \times 233 \text{ g/mol} = 2.33 \times 10^{-3} \text{ g/dm}^3$.

(A) Let solubility $S = 1 \times 10^{-3} \text{ mol/L}$.
For $MX$: $K_{sp} = S^2 = (10^{-3})^2 = 1 \times 10^{-6}$.
For $MX_2$: $K_{sp} = (S)(2S)^2 = 4S^3 = 4(10^{-3})^3 = 4 \times 10^{-9}$.
For $MX_3$: $K_{sp} = (S)(3S)^3 = 27S^4 = 27(10^{-3})^4 = 27 \times 10^{-12}$.

(B) Step $1$: Calculate the molar solubility $(S)$ of the salt.
$S = \frac{\text{moles of solute}}{\text{volume of solution in L}} = \frac{10^{-4} \text{ mol}}{0.1 \text{ L}} = 10^{-3} \text{ M}$.
Step $2$: For a binary salt $BA$, the dissociation is $BA(s) \rightleftharpoons B^+(aq) + A^-(aq)$.
Step $3$: The solubility product expression is $K_{sp} = [B^+][A^-] = S \times S = S^2$.
Step $4$: Substitute the value of $S$: $K_{sp} = (10^{-3})^2 = 10^{-6}$.

(A) The dissociation of $AgCl$ is given by: $AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)$.
Let the solubility of $AgCl$ be $S \text{ mol/L}$.
In $0.1 \text{ M } NaCl$, the concentration of $Cl^-$ ions is $0.1 \text{ M}$ from $NaCl$ and $S \text{ M}$ from $AgCl$. Since $S$ is very small, $[Cl^-] \approx 0.1 \text{ M}$.
The solubility product expression is $K_{sp} = [Ag^+][Cl^-]$.
Substituting the values: $1.8 \times 10^{-10} = (S)(0.1)$.
Solving for $S$: $S = \frac{1.8 \times 10^{-10}}{0.1} = 1.8 \times 10^{-9} \text{ M}$.

(A) Step $1$: For a salt $BA$, the solubility product $K_{sp} = S^2$, where $S$ is the molar solubility.
Step $2$: Calculate molar solubility $S = \sqrt{K_{sp}} = \sqrt{6.4 \times 10^{-13}} = \sqrt{64 \times 10^{-14}} = 8 \times 10^{-7} \text{ mol dm}^{-3}$.
Step $3$: Convert molar solubility to $\text{g dm}^{-3}$ using the molar mass $M = 190 \text{ g mol}^{-1}$.
Step $4$: Solubility in $\text{g dm}^{-3} = S \times M = (8 \times 10^{-7} \text{ mol dm}^{-3}) \times (190 \text{ g mol}^{-1}) = 1520 \times 10^{-7} \text{ g dm}^{-3} = 1.52 \times 10^{-4} \text{ g dm}^{-3}$.

(C) The dissociation of the salt is $BA(s) \rightleftharpoons B^+(aq) + A^-(aq)$.
The solubility product expression is $K_{sp} = [B^+][A^-]$.
Given $K_{sp} = 4 \times 10^{-13}$ and $[B^+] = 1 \times 10^{-6} \text{ M}$.
Substituting the values: $4 \times 10^{-13} = (1 \times 10^{-6}) \times [A^-]$.
Therefore, $[A^-] = \frac{4 \times 10^{-13}}{1 \times 10^{-6}} = 4 \times 10^{-7} \text{ M}$.

(C) The dissociation of the salt $BA$ is given by: $BA(s) \rightleftharpoons B^+(aq) + A^-(aq)$.
Let the solubility of $BA$ be $S \text{ mol dm}^{-3}$.
Then, $[B^+] = S$ and $[A^-] = S$.
The solubility product expression is $K_{sp} = [B^+][A^-] = S \times S = S^2$.
Given $K_{sp} = 8.56 \times 10^{-5}$.
Therefore, $S = \sqrt{K_{sp}} = \sqrt{8.56 \times 10^{-5}}$.
$S = \sqrt{85.6 \times 10^{-6}} \approx 9.252 \times 10^{-3} \text{ mol dm}^{-3}$.

(B) The chemical formula for silver oxalate is $Ag_2C_2O_4$.
The dissociation equilibrium is: $Ag_2C_2O_4(s) \rightleftharpoons 2Ag^+(aq) + C_2O_4^{2-}(aq)$.
Let the solubility be $S \text{ mol/L}$. Then $[Ag^+] = 2S$ and $[C_2O_4^{2-}] = S$.
The solubility product expression is $K_{sp} = [Ag^+]^2 [C_2O_4^{2-}]$.
Substituting the values: $K_{sp} = (2S)^2(S) = 4S^2 \times S = 4S^3$.
Rearranging for $S$: $S^3 = \frac{K_{sp}}{4}$, therefore $S = \sqrt[3]{\frac{K_{sp}}{4}}$.

(B) Step $1$: Calculate molar solubility $(S)$ in $\text{mol/dm}^3$ by dividing solubility in $\text{g/dm}^3$ by molar mass.
$S = \frac{18.78 \times 10^{-4} \text{ g/dm}^3}{187.8 \text{ g/mol}} = 10^{-5} \text{ mol/dm}^3$.
Step $2$: The dissociation of $AB_2$ is $AB_2(s) \rightleftharpoons A^{2+}(aq) + 2B^-(aq)$.
Step $3$: The solubility product expression is $K_{sp} = [A^{2+}][B^-]^2 = (S)(2S)^2 = 4S^3$.
Step $4$: Substitute $S = 10^{-5}$ into the expression: $K_{sp} = 4(10^{-5})^3 = 4 \times 10^{-15}$.

(D) The dissociation of the salt $AB_2$ is given by: $AB_2(s) \rightleftharpoons A^{2+}(aq) + 2B^-(aq)$.
Let the solubility be $S = 1 \times 10^{-6} \text{ mol/dm}^3$.
The concentrations are $[A^{2+}] = S$ and $[B^-] = 2S$.
The solubility product $K_{sp}$ is defined as: $K_{sp} = [A^{2+}][B^-]^2$.
Substituting the values: $K_{sp} = (S)(2S)^2 = 4S^3$.
$K_{sp} = 4 \times (1 \times 10^{-6})^3$.
$K_{sp} = 4 \times 10^{-18}$.

(D) The dissociation of the salt $AX_2$ is given by: $AX_2(s) \rightleftharpoons A^{2+}(aq) + 2X^-(aq)$.
Let the solubility be $S = 6 \times 10^{-12} \text{ mol dm}^{-3}$.
Then $[A^{2+}] = S$ and $[X^-] = 2S$.
The solubility product expression is $K_{sp} = [A^{2+}][X^-]^2$.
Substituting the values: $K_{sp} = (S)(2S)^2 = 4S^3$.
$K_{sp} = 4 \times (6 \times 10^{-12})^3$.
$K_{sp} = 4 \times (216 \times 10^{-36}) = 864 \times 10^{-36}$.
$K_{sp} = 8.64 \times 10^{-34}$.

(A) $1$. The mixture is a buffer solution consisting of a weak acid and its conjugate base.
$2$. Use the Henderson-Hasselbalch equation: $pH = pK_a + \log\left(\frac{[\text{salt}]}{[\text{acid}]}\right)$.
$3$. Since equal volumes are mixed, the final concentration of both components is halved, but their ratio remains $\frac{0.1}{0.1} = 1$.
$4$. $pK_a = -\log(K_a) = -\log(1.3 \times 10^{-5}) = 5 - \log(1.3) \approx 5 - 0.1139 = 4.8861$.
$5$. $pH = 4.8861 + \log(1) = 4.8861 \approx 4.89$.

(C) Step $1$: Identify the chemical names for the given compounds.
Step $2$: $Na_2CO_3 \cdot 10H_2O$ is Washing soda $(IV)$.
Step $3$: $NaOH$ is Caustic soda $(I)$.
Step $4$: $Ca(OH)_2$ is Slaked lime $(II)$.
Step $5$: $NaHCO_3$ is Baking soda $(III)$.
Step $6$: Therefore, the correct matching is $A-IV, B-I, C-II, D-III$.

(B) Given: $[H_3O^+] = 1 \times 10^{-4} \text{ M}$.
Step $1$: Calculate $pH$ using the formula $pH = -\log[H_3O^+]$.
$pH = -\log(10^{-4}) = 4$.
Step $2$: Use the relation $pH + pOH = 14$ at $298 \text{ K}$.
$pOH = 14 - pH = 14 - 4 = 10$.

(B) The relationship between $pH$ and $[H_3O^+]$ is given by the formula: $[H_3O^+] = 10^{-pH}$.
Substitute the given $pH$ value: $[H_3O^+] = 10^{-4.25}$.
To solve this, write the exponent as: $10^{-4.25} = 10^{0.75} \times 10^{-5}$.
Since $10^{0.75} \approx 5.623$, the concentration is $[H_3O^+] \approx 5.623 \times 10^{-5} \text{ M}$.

(A) For a weak acid, the degree of dissociation $\alpha$ is related to concentration $C$ by the relation $\alpha \approx \sqrt{\frac{K_a}{C}}$, which implies $\alpha \propto \frac{1}{\sqrt{C}}$.
Given: $\alpha_1 = 4\% = 0.04$, $C_1 = 0.05 \text{ M}$, $C_2 = 0.1 \text{ M}$.
Using the ratio: $\frac{\alpha_2}{\alpha_1} = \sqrt{\frac{C_1}{C_2}}$.
$\frac{\alpha_2}{0.04} = \sqrt{\frac{0.05}{0.1}} = \sqrt{0.5} \approx 0.707$.
$\alpha_2 = 0.04 \times 0.707 = 0.02828$.
Percent dissociation = $0.02828 \times 100 = 2.828\% \approx 2.83\%$.

(B) Step $1$: Since the monobasic acid is completely dissociated, the concentration of hydrogen ions is $[H^+] = 0.01 \text{ M} = 10^{-2} \text{ M}$.
Step $2$: Calculate the $pH$ using the formula $pH = -\log[H^+]$. Thus, $pH = -\log(10^{-2}) = 2$.
Step $3$: At $298 \text{ K}$, the relationship between $pH$ and $pOH$ is $pH + pOH = 14$.
Step $4$: Substitute the value of $pH$: $2 + pOH = 14$, which gives $pOH = 12$.

(C) Step $1$: Since $HCl$ is a strong acid, it undergoes complete dissociation in water: $HCl \rightarrow H^+ + Cl^-$.
Step $2$: The concentration of $H^+$ ions is equal to the concentration of the $HCl$ solution, so $[H^+] = 1 \text{ M}$.
Step $3$: The formula for $pH$ is $pH = -\log[H^+]$.
Step $4$: Substituting the value, $pH = -\log(1) = 0$.

(A) Step $1$: Calculate $[H^+]$ for solution $A$. Since $pH_A = 4$, $[H^+]_A = 10^{-pH_A} = 10^{-4} \text{ M}$.
Step $2$: Calculate $[H^+]$ for solution $B$. Given $[H^+]_B = 3 \times [H^+]_A = 3 \times 10^{-4} \text{ M}$.
Step $3$: Calculate $pH_B$. $pH_B = -\log[H^+]_B = -\log(3 \times 10^{-4}) = -(\log 3 + \log 10^{-4}) = -(0.4771 - 4) = 4 - 0.4771 = 3.5229$.

(A) Step $1$: Calculate the total volume of the metal sample.
Volume = $\frac{\text{mass}}{\text{density}} = \frac{18 \text{ g}}{7.2 \text{ g cm}^{-3}} = 2.5 \text{ cm}^3$.
Step $2$: Calculate the number of unit cells.
Number of unit cells = $\frac{\text{Total volume}}{\text{Volume of one unit cell}} = \frac{2.5 \text{ cm}^3}{2.5 \times 10^{-23} \text{ cm}^3} = 1.0 \times 10^{23}$.

(D) The orthorhombic crystal system is unique because it exhibits all four types of unit cells: primitive (simple), body-centered, face-centered, and end-centered. Therefore, the total number of Bravais lattices in the orthorhombic crystal system is $4$.

(C) Step $1$: Analyze the nature of the given substances.
Step $2$: $Fe$ is a metallic crystal consisting of metal atoms held by metallic bonds.
Step $3$: $CH_4$ is a molecular crystal consisting of non-polar molecules held by weak van der Waals forces.
Step $4$: $SiO_2$ (quartz) is a covalent network crystal where $Si$ and $O$ atoms are linked by a continuous network of covalent bonds.
Step $5$: $KCl$ is an ionic crystal consisting of $K^+$ and $Cl^-$ ions held by strong electrostatic forces.
Step $6$: Therefore, $SiO_2$ is the correct covalent crystal.

(A) Osmotic pressure $\pi = CRT$, where $C$ is molarity. For the same temperature $T$, solutions with the same molarity $C$ have the same osmotic pressure.
Step $1$: Calculate molarity $C = \frac{\text{mass (g/L)}}{\text{molar mass (g/mol)}}$.
Step $2$: For urea $(M = 60 \text{ g/mol})$, $C_1 = \frac{6 \text{ g/L}}{60 \text{ g/mol}} = 0.1 \text{ M}$.
Step $3$: For glucose $(M = 180 \text{ g/mol})$, $C_2 = \frac{18 \text{ g/L}}{180 \text{ g/mol}} = 0.1 \text{ M}$.
Step $4$: Since $C_1 = C_2$, the osmotic pressures are equal.

(B) $1$. Elevation in boiling point is given by $\Delta T_b = T_b - T_b^\circ = 357.27 \text{ K} - 350.06 \text{ K} = 7.21 \text{ K}$.
$2$. The formula for elevation in boiling point is $\Delta T_b = K_b \times m$, where $m$ is molality.
$3$. Molality $m = \frac{W_2 \times 1000}{M_2 \times W_1}$, where $W_2 = 50 \text{ g}$, $W_1 = 150 \text{ g}$, and $M_2$ is the molar mass of solute.
$4$. Substituting values: $7.21 = 2.77 \times \frac{50 \times 1000}{M_2 \times 150}$.
$5$. $M_2 = \frac{2.77 \times 50000}{7.21 \times 150} = \frac{138500}{1081.5} \approx 128.06 \text{ g/mol}$.
$6$. Thus, the molar mass is approximately $128 \text{ g/mol}$.

(D) $1$. Vapour pressure of a solution decreases as the concentration of solute particles increases (colligative property).
$2$. The number of particles produced by dissociation in $1 \text{ M}$ solutions are:
- $1 \text{ M NaCl} \rightarrow 1 \text{ M } Na^+ + 1 \text{ M } Cl^- = 2 \text{ M particles}$.
- $1 \text{ M CaCl}_2 \rightarrow 1 \text{ M } Ca^{2+} + 2 \text{ M } Cl^- = 3 \text{ M particles}$.
- $1 \text{ M AlCl}_3 \rightarrow 1 \text{ M } Al^{3+} + 3 \text{ M } Cl^- = 4 \text{ M particles}$.
- $1 \text{ M C}_{12}\text{H}_{22}\text{O}_{11}$ (sucrose) does not dissociate, so it remains $1 \text{ M particle}$.
$3$. Since sucrose has the minimum number of particles, it will cause the minimum lowering of vapour pressure, resulting in the maximum vapour pressure among the given options.

(C) Given:
Mass of solute $(w_2)$ = $2 \text{ g}$
Mass of solvent $(w_1)$ = $50 \text{ g} = 0.05 \text{ kg}$
Elevation in boiling point $(\Delta T_b)$ = $0.13 \text{ K}$
Molal elevation constant $(K_b)$ = $0.52 \text{ K kg mol}^{-1}$
Formula: $\Delta T_b = K_b \times m = K_b \times \frac{w_2 \times 1000}{M_2 \times w_1 \text{ (in g)}}$
Rearranging for molar mass $(M_2)$: $M_2 = \frac{K_b \times w_2 \times 1000}{\Delta T_b \times w_1}$
$M_2 = \frac{0.52 \times 2 \times 1000}{0.13 \times 50}$
$M_2 = \frac{1040}{6.5} = 160 \text{ g mol}^{-1}$

(B) The formula for osmotic pressure $(\pi)$ is given by: $\pi = CRT$
Where:
$\pi = 4.92 \text{ atm}$
$R = 0.082 \text{ L atm mol}^{-1} \text{ K}^{-1}$
$T = 300 \text{ K}$
Rearranging the formula to solve for molarity $(C)$:
$C = \frac{\pi}{RT}$
$C = \frac{4.92}{0.082 \times 300}$
$C = \frac{4.92}{24.6}$
$C = 0.2 \text{ M}$
Thus, the molarity of the solution is $0.2 \text{ M}$.

(B) The van't Hoff factor $i$ is defined as the ratio of theoretical molar mass to experimentally observed molar mass: $i = \frac{M_{\text{theoretical}}}{M_{\text{observed}}}$.
Given $i = 3$.
For $KCl \rightarrow K^+ + Cl^-$, $i = 2$.
For $K_2SO_4 \rightarrow 2K^+ + SO_4^{2-}$, $i = 3$.
For $MgSO_4 \rightarrow Mg^{2+} + SO_4^{2-}$, $i = 2$.
For $Al_2(SO_4)_3 \rightarrow 2Al^{3+} + 3SO_4^{2-}$, $i = 5$.
Thus, $K_2SO_4$ has $i = 3$.

(C) Step $1$: Calculate the molality of the urea solution.
Molality $(m) = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}$.
Moles of urea $= \frac{5 \text{ g}}{60 \text{ g mol}^{-1}} = \frac{1}{12} \text{ mol}$.
Molality of urea solution $(m_1) = \frac{1/12 \text{ mol}}{0.1 \text{ kg}} = \frac{1}{1.2} \text{ mol kg}^{-1}$.
Step $2$: Since boiling points are the same, the molality of the glucose solution must be equal to the molality of the urea solution $(m_2 = m_1)$.
Let $x$ be the mass of glucose.
Moles of glucose $= \frac{x}{180} \text{ mol}$.
Molality of glucose solution $(m_2) = \frac{x/180 \text{ mol}}{0.12 \text{ kg}} = \frac{x}{180 \times 0.12} \text{ mol kg}^{-1} = \frac{x}{21.6} \text{ mol kg}^{-1}$.
Step $3$: Equate the molalities:
$\frac{1}{1.2} = \frac{x}{21.6}$.
$x = \frac{21.6}{1.2} = 18 \text{ g}$.
Therefore, the required amount of glucose is $18 \text{ g}$.

(A) Step $1$: Calculate the number of moles of solute $(n)$: $n = \frac{\text{mass}}{\text{molar mass}} = \frac{40 \text{ g}}{180 \text{ g mol}^{-1}} = \frac{2}{9} \text{ mol} \approx 0.222 \text{ mol}$.
Step $2$: Use the osmotic pressure formula $\Pi V = nRT$, where $\Pi = 2 \text{ atm}$, $n = \frac{2}{9} \text{ mol}$, $R = 0.0821 \text{ L atm K}^{-1} \text{mol}^{-1}$, and $T = 300 \text{ K}$.
Step $3$: Rearrange to solve for $V$: $V = \frac{nRT}{\Pi} = \frac{(2/9) \times 0.0821 \times 300}{2}$.
Step $4$: Calculate the value: $V = \frac{0.2222 \times 0.0821 \times 300}{2} = \frac{5.472}{2} = 2.736 \text{ L}$.
Note: Since $1 \text{ L} = 1 \text{ dm}^3$, the volume is approximately $2.74 \text{ dm}^3$. Given the options, there appears to be a discrepancy in the provided choices. Re-evaluating with $R \approx 0.082$ and $n = 40/180$: $V = (40/180 \times 0.0821 \times 300) / 2 = 2.736 \text{ dm}^3$. None of the options match.

(C) The formula for osmotic pressure $(\pi)$ is $\pi = CRT = \frac{n}{V} RT = \frac{w}{M \cdot V} RT$.
Given: $\pi = 2.4 \text{ atm}$, $w = 6 \text{ g}$, $V = 1 \text{ dm}^3$, $T = 300 \text{ K}$, $R = 0.0821 \text{ atm dm}^3 \text{ K}^{-1} \text{ mol}^{-1}$.
Rearranging for molar mass $(M)$: $M = \frac{wRT}{\pi V}$.
Substituting the values: $M = \frac{6 \times 0.0821 \times 300}{2.4 \times 1}$.
$M = \frac{147.78}{2.4} = 61.575 \text{ g mol}^{-1} \approx 61.58 \text{ g mol}^{-1}$.

(C) The dissociation reaction for $BaCl_2$ is: $BaCl_2 \rightarrow Ba^{2+} + 2Cl^-$.
The number of ions produced per formula unit is $n = 3$.
The relationship between van't Hoff factor $(i)$ and degree of dissociation $(\alpha)$ is given by: $i = 1 + (n - 1)\alpha$.
Given $i = 2.47$ and $n = 3$:
$2.47 = 1 + (3 - 1)\alpha$
$2.47 = 1 + 2\alpha$
$1.47 = 2\alpha$
$\alpha = \frac{1.47}{2} = 0.735$.
Percentage dissociation = $\alpha \times 100 = 0.735 \times 100 = 73.5\%$.

(C) The formula for boiling point elevation is $\Delta T_b = K_b \times m$, where $\Delta T_b$ is the boiling point elevation, $K_b$ is the molal boiling point elevation constant, and $m$ is the molality.
Given: $\Delta T_b = 0.70 \text{ K}$, $K_b = 2.44 \text{ K kg mol}^{-1}$.
Rearranging the formula to solve for molality: $m = \frac{\Delta T_b}{K_b}$.
Substituting the values: $m = \frac{0.70 \text{ K}}{2.44 \text{ K kg mol}^{-1}} \approx 0.2868 \text{ mol kg}^{-1}$.
Rounding to three decimal places, $m \approx 0.287 \text{ m}$. The closest option is $0.286 \text{ m}$.

(D) Step $1$: Calculate the depression in freezing point $(\Delta T_f)$.
$\Delta T_f = T_f^\circ - T_f = 0 \text{ }^\circ\text{C} - (-0.93 \text{ }^\circ\text{C}) = 0.93 \text{ K}$.
Step $2$: Use the formula for depression in freezing point with the van't Hoff factor:
$\Delta T_f = i \times K_f \times m$.
Step $3$: Rearrange the formula to solve for molality $(m)$:
$m = \frac{\Delta T_f}{i \times K_f}$.
Step $4$: Substitute the given values:
$m = \frac{0.93}{1.25 \times 1.86} = \frac{0.93}{2.325} = 0.4 \text{ m}$.
Thus, the molality of the solution is $0.4 \text{ m}$.

(C) The freezing point depression is given by $\Delta T_f = i \cdot K_f \cdot m$, where $i$ is the van't Hoff factor and $m$ is the molality.
For complete dissociation, $i$ equals the number of ions produced per formula unit.
$(a)$ $KCl \rightarrow K^+ + Cl^-$, $i = 2$. $\Delta T_f = 2 \times 0.2 = 0.4 \text{ m}$.
$(b)$ $NaCl \rightarrow Na^+ + Cl^-$, $i = 2$. $\Delta T_f = 2 \times 0.1 = 0.2 \text{ m}$.
$(c)$ $AlPO_4 \rightarrow Al^{3+} + PO_4^{3-}$, $i = 2$. $\Delta T_f = 2 \times 0.05 = 0.1 \text{ m}$.
$(d)$ $MgSO_4 \rightarrow Mg^{2+} + SO_4^{2-}$, $i = 2$. $\Delta T_f = 2 \times 0.15 = 0.3 \text{ m}$.
Comparing the values, $0.1 \text{ m}$ is the minimum value, which corresponds to $0.05 \text{ m Aluminium phosphate}$.

(A) The formula for osmotic pressure is $\pi = CRT$, where $C = \frac{n}{V} = \frac{w}{M \times V}$.
Given: $\pi = 0.1 \text{ atm}$, $V = 0.3 \text{ dm}^3$, $T = 300 \text{ K}$, $M = 328 \text{ g mol}^{-1}$, $R = 0.082 \text{ dm}^3 \text{atm K}^{-1} \text{mol}^{-1}$.
Rearranging the formula for mass $w$: $w = \frac{\pi \times M \times V}{R \times T}$.
Substituting the values: $w = \frac{0.1 \times 328 \times 0.3}{0.082 \times 300}$.
$w = \frac{9.84}{24.6} = 0.4 \text{ g}$.

(A) The formula for elevation in boiling point is $\Delta T_b = K_b \times m$.
Given:
Molal elevation constant, $K_b = 3.0 \text{ K kg mol}^{-1}$.
Molality of the solution, $m = 0.45 \text{ m}$.
Substituting the values into the formula:
$\Delta T_b = 3.0 \text{ K kg mol}^{-1} \times 0.45 \text{ mol kg}^{-1} = 1.35 \text{ K}$.
Thus, the elevation in boiling point is $1.35 \text{ K}$.

(B) Colligative properties are those properties of a solution that depend only on the number of solute particles present in the solution, not on their identity.
The four main colligative properties are:
$1$. Relative lowering of vapour pressure.
$2$. Elevation in boiling point.
$3$. Depression in freezing point.
$4$. Osmotic pressure.
Among the given options, only 'Osmotic pressure' is a colligative property. Vapour pressure, boiling point, and freezing point are properties of the solution, but their changes (lowering, elevation, and depression respectively) are the colligative properties.

(B) The formula for osmotic pressure $(\pi)$ is $\pi = CRT$.
Given:
Concentration $C = 0.01 \text{ M} = 0.01 \text{ mol L}^{-1}$.
Temperature $T = 25 \text{ }^\circ\text{C} = 25 + 273 = 298 \text{ K}$.
Gas constant $R = 0.0821 \text{ L atm mol}^{-1} \text{K}^{-1}$.
Substituting the values:
$\pi = 0.01 \text{ mol L}^{-1} \times 0.0821 \text{ L atm mol}^{-1} \text{K}^{-1} \times 298 \text{ K}$.
$\pi = 0.01 \times 24.4658 \text{ atm}$.
$\pi = 0.244658 \text{ atm} \approx 0.245 \text{ atm}$.

(A) According to Henry's Law, the solubility $(S)$ of a gas is directly proportional to its partial pressure $(P)$: $S = k_H \times P$.
Step $1$: Calculate Henry's constant $(k_H)$ using the initial conditions:
$k_H = \frac{S_1}{P_1} = \frac{6.85 \times 10^{-4} \text{ mol L}^{-1}}{1 \text{ bar}} = 6.85 \times 10^{-4} \text{ mol L}^{-1} \text{ bar}^{-1}$.
Step $2$: Calculate the new solubility $(S_2)$ at $P_2 = 0.70 \text{ bar}$:
$S_2 = k_H \times P_2 = (6.85 \times 10^{-4} \text{ mol L}^{-1} \text{ bar}^{-1}) \times (0.70 \text{ bar}) = 4.795 \times 10^{-4} \text{ mol L}^{-1} \approx 4.80 \times 10^{-4} \text{ mol L}^{-1}$.

(A) According to Henry's Law, the concentration of a dissolved gas is given by the formula: $C = K_H \times P$
Given:
Partial pressure, $P = 0.18 \text{ atm}$
Henry's Law constant, $K_H = 0.15 \text{ mol dm}^{-3} \text{atm}^{-1}$
Calculation:
$C = 0.15 \text{ mol dm}^{-3} \text{atm}^{-1} \times 0.18 \text{ atm}$
$C = 0.027 \text{ mol dm}^{-3}$
Since $1 \text{ mol dm}^{-3} = 1 \text{ M}$, the concentration is $0.027 \text{ M}$.

(A) Henry's law is applicable to gases that do not undergo any chemical reaction with the solvent. $NH_3$ and $CO_2$ react with water to form $NH_4OH$ and $H_2CO_3$ respectively. Therefore, these gases do not obey Henry's law.

(C) According to Henry's law, the solubility $(S)$ of a gas is given by the formula: $S = K_H \times P$
Given:
$K_H = 3.0 \times 10^{-2} \text{ mol dm}^{-3} \text{ atm}^{-1}$
$P = 3 \text{ atm}$
Substituting the values:
$S = (3.0 \times 10^{-2} \text{ mol dm}^{-3} \text{ atm}^{-1}) \times (3 \text{ atm})$
$S = 9.0 \times 10^{-2} \text{ mol dm}^{-3}$
$S = 0.09 \text{ M}$
Therefore, the solubility is $0.09 \text{ M}$.

(A) $1$. Soft drinks are bottled under high pressure to increase the solubility of $CO_2$ gas in the liquid (Henry's Law).
$2$. When the bottle is opened, the internal pressure drops to atmospheric pressure.
$3$. Due to the decrease in pressure, the solubility of the dissolved $CO_2$ gas decreases (Statement $A$ is correct).
$4$. The bottle releases the excess internal pressure (Statement $B$ is correct).
$5$. The dissolved gas escapes rapidly, causing effervescence (Statement $C$ is correct).
$6$. External pressure remains constant; it does not increase (Statement $D$ is incorrect).
$7$. Therefore, $(A)$, $(B)$, and $(C)$ are correct.

(A) $1$. Calculate mass of $CCl_4$: $\text{Mass} = \text{Density} \times \text{Volume} = 1.58 \text{ g/cm}^3 \times 100 \text{ cm}^3 = 158 \text{ g}$.
$2$. Calculate moles of $CCl_4$ $(M_{CCl_4} = 12 + 4 \times 35.5 = 154 \text{ g/mol})$: $n_{CCl_4} = \frac{158}{154} \approx 1.026 \text{ mol}$.
$3$. Calculate moles of solute: $n_{\text{solute}} = \frac{0.5}{65} \approx 0.00769 \text{ mol}$.
$4$. Calculate mole fraction of solvent $(X_A)$: $X_A = \frac{n_{CCl_4}}{n_{CCl_4} + n_{\text{solute}}} = \frac{1.026}{1.026 + 0.00769} = \frac{1.026}{1.03369} \approx 0.99256$.
$5$. Calculate vapour pressure of solution $(P_s)$: $P_s = P^0 \times X_A = 143 \text{ mm Hg} \times 0.99256 \approx 141.93 \text{ mm Hg}$.

(A) Step $1$: Identify the formula for vapour pressure of a solution using Raoult's Law: $P_s = P^0 \times x_1$, where $P_s$ is the vapour pressure of the solution, $P^0$ is the vapour pressure of pure solvent, and $x_1$ is the mole fraction of the solvent.
Step $2$: Substitute the given values into the formula: $P^0 = 18 \text{ mm Hg}$ and $x_1 = 0.9964$.
Step $3$: Calculate the result: $P_s = 18 \text{ mm Hg} \times 0.9964 = 17.9352 \text{ mm Hg}$.
Step $4$: Rounding to two decimal places, we get $P_s \approx 17.94 \text{ mm Hg}$.

(D) According to Raoult's law, the relative lowering of vapour pressure is equal to the mole fraction of the solute.
$\frac{P^o - P_s}{P^o} = X_{\text{solute}}$
Given that the vapour pressure decreases by $20 \%$, the lowering of vapour pressure is $\frac{P^o - P_s}{P^o} = 0.2$.
Therefore, the mole fraction of the solute $X_{\text{solute}} = 0.2$.
Since the sum of mole fractions in a binary solution is $1$, we have $X_{\text{solvent}} + X_{\text{solute}} = 1$.
$X_{\text{solvent}} = 1 - 0.2 = 0.8$.

(B) Given:
Vapour pressure of pure solvent, $P^\circ_A = 30 \text{ mm Hg}$
Vapour pressure of solution, $P_s = 27 \text{ mm Hg}$
According to Raoult's law for a solution containing a non-volatile solute:
$\frac{P^\circ_A - P_s}{P^\circ_A} = x_B$
where $x_B$ is the mole fraction of the solute.
Substituting the values:
$x_B = \frac{30 - 27}{30}$
$x_B = \frac{3}{30}$
$x_B = 0.1$
Thus, the mole fraction of the solute is $0.1$.

(B) $1$. $A$ solution is a homogeneous mixture of two or more substances.
$2$. In a solution, the component present in a smaller amount is the solute, and the component present in a larger amount is the solvent.
$3$. Sea water is a solution where salts (solids) are dissolved in water (liquid).
$4$. Hydrogen in Palladium is a gas in solid solution.
$5$. Gels are colloids where a liquid is dispersed in a solid.
$6$. Amalgams are solutions of metals (solids) in mercury (liquid), but in common classification, sea water is the standard example of a solid solute in a liquid solvent.

(D) Step $1$: Henry's law states that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid.
Step $2$: However, Henry's law is only applicable to gases that do not undergo chemical reactions with the solvent.
Step $3$: Gases like $NH_3$ and $CO_2$ react with water (e.g., $NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^-$), therefore they do not strictly obey Henry's law.
Step $4$: Thus, statement $(D)$ is not true.

(A) Brass is an alloy of copper and zinc. Since both copper and zinc are solids at room temperature, brass is a homogeneous mixture of two solids. Therefore, it is classified as a solid-in-solid solution.

(A) The formula for molar conductivity $(\Lambda_m)$ is given by: $\Lambda_m = \frac{\kappa \times 1000}{M}$
Given: $\kappa = 2.50 \times 10^{-4} \text{ S cm}^{-1}$ and $M = 0.0020 \text{ mol L}^{-1}$.
Substituting the values: $\Lambda_m = \frac{2.50 \times 10^{-4} \times 1000}{0.0020}$
$\Lambda_m = \frac{0.25}{0.0020} = \frac{2500}{20} = 125 \text{ S cm}^2 \text{mol}^{-1}$.

(C) $1$. Molar conductivity $(\Lambda_m)$ is defined as $\Lambda_m = \frac{\kappa}{C}$. Upon dilution, the concentration $(C)$ decreases, which leads to an increase in the volume containing $1 \text{ mole}$ of electrolyte, thereby increasing $\Lambda_m$.
$2$. Conductivity $(\kappa)$ is defined as the conductance of $1 \text{ cm}^3$ of solution. Upon dilution, the number of ions per unit volume decreases, which results in a decrease in $\kappa$.
$3$. Therefore, $\Lambda_m$ increases and $\kappa$ decreases.

(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, each ion makes a definite contribution to the molar conductivity of an electrolyte, which is independent of the presence of the other ion.
Step $3$: Therefore, the correct law is Kohlrausch's law of independent migration of ions.

(C) The formula for molar conductivity $(\Lambda_m)$ is given by: $\Lambda_m = \frac{\kappa \times 1000}{C}$.
Given: Conductivity $\kappa = 0.0248 \text{ }\Omega^{-1} \text{cm}^{-1}$ and Concentration $C = 0.2 \text{ M}$.
Substituting the values: $\Lambda_m = \frac{0.0248 \times 1000}{0.2}$.
$\Lambda_m = \frac{24.8}{0.2} = 124 \text{ }\Omega^{-1} \text{cm}^2 \text{mol}^{-1}$.

(C) $1$. Electrical conductivity in a solution depends on the presence of free ions.
$2$. $Distilled \ water$ is a poor conductor as it lacks ions.
$3$. $Sugar$ and $Alcohol$ are covalent compounds that do not dissociate into ions in aqueous solution.
$4$. $Salt$ $(NaCl)$ is an ionic compound that dissociates completely into $Na^+$ and $Cl^-$ ions in water, making it a strong electrolyte with high electrical conductivity.

(B) The relationship between molar conductivity $(\Lambda_m)$ and conductivity $(\kappa)$ is given by the formula: $\Lambda_m = \frac{\kappa \times 1000}{M}$, where $M$ is the molarity.
Given: $\Lambda_m = 200 \text{ }\Omega^{-1} \text{cm}^2 \text{mol}^{-1}$ and $M = 0.04 \text{ M}$.
Rearranging the formula to solve for $\kappa$: $\kappa = \frac{\Lambda_m \times M}{1000}$.
Substituting the values: $\kappa = \frac{200 \times 0.04}{1000}$.
$\kappa = \frac{8}{1000} = 0.008 \text{ }\Omega^{-1} \text{cm}^{-1}$.

(B) The relationship between molar conductivity $(\Lambda_m)$, conductivity $(\kappa)$, and molarity $(C)$ is given by the formula: $\Lambda_m = \frac{\kappa \times 1000}{C}$.
Given: $\Lambda_m = 120 \text{ ohm}^{-1} \text{cm}^2 \text{mol}^{-1}$, $\kappa = 0.0024 \text{ ohm}^{-1} \text{cm}^{-1}$.
Rearranging the formula for concentration $(C)$: $C = \frac{\kappa \times 1000}{\Lambda_m}$.
Substituting the values: $C = \frac{0.0024 \times 1000}{120}$.
$C = \frac{2.4}{120} = 0.02 \text{ M}$.

(A) The relationship between conductivity $(\kappa)$, resistance $(R)$, and cell constant $(G^*)$ is given by the formula: $\kappa = \frac{G^*}{R}$.
Rearranging for the cell constant: $G^* = \kappa \times R$.
Given: $\kappa = 0.0123 \text{ ohm}^{-1} \text{ cm}^{-1}$ and $R = 120 \text{ ohm}$.
Substituting the values: $G^* = 0.0123 \text{ ohm}^{-1} \text{ cm}^{-1} \times 120 \text{ ohm} = 1.476 \text{ cm}^{-1}$.

(D) $1$. According to the Kohlrausch law, for strong electrolytes, the variation of molar conductivity $(\Lambda_m)$ with concentration $(c)$ is given by the Debye-$H$ückel-Onsager equation: $\Lambda_m = \Lambda_m^0 - A\sqrt{c}$.
$2$. This equation represents a straight line with a negative slope and an intercept equal to $\Lambda_m^0$ on the y-axis.
$3$. Weak electrolytes like $CH_3COOH$, $NH_4OH$, and $HCOOH$ show a non-linear increase in $\Lambda_m$ with dilution, especially at low concentrations.
$4$. $CH_3COONa$ is a strong electrolyte, which follows the linear relationship.
$5$. Therefore, '$X$' is $CH_3COONa$.

(A) The formula for conductivity $(\kappa)$ is given by: $\kappa = \frac{\text{Cell constant}}{\text{Resistance}}$
Given: Cell constant $(G^*)$ = $1.2 \text{ cm}^{-1}$, Resistance $(R)$ = $95.0 \text{ }\Omega$.
Substituting the values: $\kappa = \frac{1.2 \text{ cm}^{-1}}{95.0 \text{ }\Omega} = 0.01263 \text{ }\Omega^{-1} \text{ cm}^{-1}$.
Expressing in scientific notation: $\kappa = 1.26 \times 10^{-2} \text{ }\Omega^{-1} \text{ cm}^{-1}$.

(C) The relationship between molar conductivity $(\Lambda_m)$, conductivity $(\kappa)$, and molarity $(M)$ is given by: $\Lambda_m = \frac{\kappa \times 1000}{M}$.
Given: $\Lambda_m = 240 \text{ }\Omega^{-1}\text{cm}^2\text{mol}^{-1}$, $\kappa = 0.012 \text{ }\Omega^{-1}\text{cm}^{-1}$.
Rearranging the formula for $M$: $M = \frac{\kappa \times 1000}{\Lambda_m}$.
Substituting the values: $M = \frac{0.012 \times 1000}{240}$.
$M = \frac{12}{240} = \frac{1}{20} = 0.05 \text{ M}$.

(A) According to Kohlrausch's law, the molar conductivity at infinite dilution $(\Lambda^0_m)$ can be expressed as the sum of the ionic conductivities.
Given:
$\Lambda^0_m(H_2SO_4) = 2\lambda^0(H^+) + \lambda^0(SO_4^{2-}) = x$
$\Lambda^0_m(K_2SO_4) = 2\lambda^0(K^+) + \lambda^0(SO_4^{2-}) = y$
$\Lambda^0_m(CH_3COOK) = \lambda^0(CH_3COO^-) + \lambda^0(K^+) = z$
We need to find $\Lambda^0_m(CH_3COOH) = \lambda^0(H^+) + \lambda^0(CH_3COO^-)$.
From the given equations:
$\frac{\Lambda^0_m(H_2SO_4) - \Lambda^0_m(K_2SO_4)}{2} = \frac{(2\lambda^0(H^+) + \lambda^0(SO_4^{2-})) - (2\lambda^0(K^+) + \lambda^0(SO_4^{2-}))}{2} = \lambda^0(H^+) - \lambda^0(K^+) = \frac{x - y}{2}$.
Adding this to $\Lambda^0_m(CH_3COOK)$:
$\frac{x - y}{2} + z = (\lambda^0(H^+) - \lambda^0(K^+)) + (\lambda^0(CH_3COO^-) + \lambda^0(K^+)) = \lambda^0(H^+) + \lambda^0(CH_3COO^-) = \Lambda^0_m(CH_3COOH)$.
Thus, the correct option is $A$.

(B) The formula for molar conductivity $(\Lambda_m)$ is given by: $\Lambda_m = \frac{\kappa \times 1000}{C}$
Given:
Conductivity $(\kappa)$ = $4.37 \times 10^{-4} \text{ }\Omega^{-1} \text{cm}^{-1}$
Concentration $(C)$ = $0.02 \text{ M}$
Substituting the values:
$\Lambda_m = \frac{4.37 \times 10^{-4} \times 1000}{0.02}$
$\Lambda_m = \frac{0.437}{0.02}$
$\Lambda_m = 21.85 \text{ }\Omega^{-1} \text{cm}^2 \text{mol}^{-1}$
Thus, the correct option is $B$.

(B) The variation of molar conductivity $(\Lambda_m)$ with concentration $(c)$ for strong electrolytes is given by the Kohlrausch equation:
$\Lambda_m = \Lambda_m^0 - A\sqrt{c}$
Where:
$1$. $\Lambda_m$ is the molar conductivity at a given concentration.
$2$. $\Lambda_m^0$ is the molar conductivity at infinite dilution.
$3$. $A$ is a constant that depends on the nature of the solvent and temperature.
$4$. $c$ is the concentration of the electrolyte.
Thus, option $B$ is correct.

(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, statement $(A)$ is correct.

(C) Step $1$: $A$ mercury battery is a primary cell that provides a constant voltage.
Step $2$: It uses a $Zn-Hg$ amalgam as the anode and $HgO$ mixed with carbon as the cathode.
Step $3$: The electrolyte used is a paste of $KOH$ and $ZnO$, which is strongly alkaline, not acidic.
Step $4$: 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 $5$: Since the electrolyte is alkaline, statement $(c)$ is incorrect.

(C) In a dry cell, also known as a Leclanché cell, the electrolyte is a moist paste of $NH_4Cl$ (ammonium chloride) and $ZnCl_2$ (zinc chloride).
$MnO_2$ (manganese dioxide) acts as the depolarizer, while $KOH$ is used in alkaline batteries, not standard dry cells.

(C) The chemical reactions occurring during the discharge of a lead storage battery are:
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 during the discharge process.