MHT CET 2026 Chemistry Question Paper with Answer and Solution

(B) $1$. In an aqueous solution, the product of $[H_3O^+]$ and $[OH^-]$ is constant at a given temperature $(K_w = [H_3O^+][OH^-])$.
$2$. $A$ solution is basic if the concentration of hydroxide ions is greater than the concentration of hydronium ions, i.e., $[OH^-] > [H_3O^+]$.
$3$. Therefore, a basic solution contains an excess of $[OH^-]$ ions.
$4$. Option $(A)$ is true for neutral solutions. Option $(C)$ is true for acidic solutions. Option $(D)$ is incorrect because $pOH = -\log[OH^-]$.

(B) The ionic product of water at $298 \text{ K}$ is given by $[H^+][OH^-] = 1.0 \times 10^{-14} \text{ M}^2$.
Given $[OH^-] = 0.08 \text{ mol/dm}^3 = 8 \times 10^{-2} \text{ M}$.
$[H^+] = \frac{1.0 \times 10^{-14}}{[OH^-]} = \frac{1.0 \times 10^{-14}}{8 \times 10^{-2}}$.
$[H^+] = 0.125 \times 10^{-12} \text{ M}$.

(B) For a weak base $BOH$, the dissociation equilibrium is $BOH \rightleftharpoons B^+ + OH^-$.
Let $C$ be the initial concentration and $\alpha$ be the degree of dissociation.
The equilibrium concentrations are $[BOH] = C(1-\alpha)$, $[B^+] = C\alpha$, and $[OH^-] = C\alpha$.
The dissociation constant $K_b$ is given by $K_b = \frac{[B^+][OH^-]}{[BOH]} = \frac{(C\alpha)(C\alpha)}{C(1-\alpha)} = \frac{C\alpha^2}{1-\alpha}$.
Since the base is weak, $\alpha \ll 1$, so $1-\alpha \approx 1$.
Thus, $K_b \approx C\alpha^2$, which gives $\alpha = \sqrt{K_b/C}$.
Therefore, $\alpha \propto \frac{1}{\sqrt{C}}$.

(B) According to the Bronsted-Lowry theory, the conjugate base of a weak acid is a strong base.
$1$. The corresponding acids for the given conjugate bases are: $HCl$ (for $Cl^-$), $CH_3COOH$ (for $CH_3COO^-$), $HNO_3$ (for $NO_3^-$), and $H_2SO_4$ (for $HSO_4^-$).
$2$. Among these acids, $CH_3COOH$ is the weakest acid.
$3$. Since the strength of a conjugate base is inversely proportional to the strength of its parent acid, $CH_3COO^-$ is the strongest base among the given options.

(C) conjugate base is formed by the removal of a proton $(H^+)$ from an acid.
For the complex ion $[Zn(H_2O)_4]^{2+}$, removing one $H^+$ ion from one of the coordinated water molecules results in:
$[Zn(H_2O)_4]^{2+} - H^+ \rightarrow [Zn(H_2O)_3(OH)]^+ + H^+$.
Therefore, the conjugate base is $[Zn(H_2O)_3(OH)]^+$.

(A) Step $1$: Strong electrolytes dissociate completely into ions in an aqueous medium, whereas weak electrolytes dissociate only partially.
Step $2$: $NaOH$, $KOH$, and $HI$ are strong electrolytes because they dissociate completely in water.
Step $3$: $NH_3$ is a weak base that undergoes partial dissociation in water to form $NH_4^+$ and $OH^-$ ions.
Step $4$: Therefore, $NH_3$ is a weak electrolyte.

(C) conjugate base is formed when an acid loses a proton $(H^+)$.
For $H_3PO_3$: $H_3PO_3 \rightarrow H^+ + H_2PO_3^-$.
For $H_2SO_4$: $H_2SO_4 \rightarrow H^+ + HSO_4^-$.
Therefore, the conjugate bases are $H_2PO_3^-$ and $HSO_4^-$ respectively.

(C) $1$. $A$ Lewis acid is defined as an electron pair acceptor.
$2$. In $BeCl_2$, the central atom $Be$ has only $4$ electrons in its valence shell, which is less than the octet ($8$ electrons).
$3$. Due to this electron deficiency, $BeCl_2$ can accept an electron pair to complete its octet, thus acting as a Lewis acid.

(B) Given: Degree of dissociation $\alpha = 1.3\% = 0.013$.
Dissociation constant $K_b = 1.69 \times 10^{-5}$.
For a weak base, the relationship between $K_b$, concentration $C$, and $\alpha$ is given by $K_b = C\alpha^2$.
Rearranging for $C$: $C = \frac{K_b}{\alpha^2}$.
Substituting the values: $C = \frac{1.69 \times 10^{-5}}{(0.013)^2}$.
$C = \frac{1.69 \times 10^{-5}}{1.69 \times 10^{-4}}$.
$C = 0.1 \text{ M}$.

(C) $1$. Electrical conductivity in a solution depends on the concentration of free ions.
$2$. $CH_3COOH$ and $NH_4OH$ are weak electrolytes that dissociate only partially.
$3$. $C_6H_{12}O_6$ (glucose) is a non-electrolyte and does not dissociate into ions.
$4$. $HCl$ is a strong acid that dissociates completely into $H^+$ and $Cl^-$ ions in water.
$5$. Since $HCl$ provides the highest concentration of ions, it is the best conductor of electricity.

(A) For a weak monobasic acid, the relationship between dissociation constant $K_a$, molar concentration $C$, and degree of dissociation $\alpha$ is given by $K_a = C\alpha^2$.
Rearranging for concentration: $C = \frac{K_a}{\alpha^2}$.
Substitute the given values: $C = \frac{1.8 \times 10^{-5}}{(0.01)^2}$.
$C = \frac{1.8 \times 10^{-5}}{10^{-4}}$.
$C = 1.8 \times 10^{-1} \text{ M} = 0.18 \text{ M}$.

(B) Let the oxidation number of $Cl$ be $x$.
For $ClO_4^-$, the sum of oxidation numbers of all atoms equals the charge on the ion.
$x + 4 \times (-2) = -1$
$x - 8 = -1$
$x = -1 + 8$
$x = +7$
Therefore, the oxidation number of $Cl$ is $+7$.

(B) Step $1$: In $OF_2$, fluorine is more electronegative than oxygen. Let the oxidation state of $O$ be $x$. Since $F$ is $-1$, we have $x + 2(-1) = 0$, which gives $x = +2$.
Step $2$: In $KO_2$ (potassium superoxide), potassium is $+1$. Let the oxidation state of $O$ be $y$. We have $1 + 2y = 0$, which gives $2y = -1$, so $y = -1/2$.

(B) Let the oxidation number of $O$ be $x$.
In $KO_2$, the oxidation number of $K$ is $+1$.
The sum of oxidation numbers in a neutral molecule is $0$.
$1 + 2x = 0$
$2x = -1$
$x = -1/2$
Thus, the oxidation number of oxygen in $KO_2$ is $-1/2$.

(C) The chemical formula for chlorous acid is $HClO_2$.
Let the oxidation state of chlorine be $x$.
The oxidation state of hydrogen is $+1$ and oxygen is $-2$.
Sum of oxidation states in a neutral molecule is $0$.
$1 + x + 2(-2) = 0$
$1 + x - 4 = 0$
$x - 3 = 0$
$x = +3$.

(B) For $SO_2$: Let the oxidation number of $S$ be $x$. Since the oxidation number of $O$ is $-2$, we have $x + 2(-2) = 0$, which gives $x = +4$.
For $SO_4^{2-}$: Let the oxidation number of $S$ be $x$. Since the oxidation number of $O$ is $-2$, we have $x + 4(-2) = -2$, which gives $x - 8 = -2$, so $x = +6$.

(B) The chemical formula of oxalic acid is $H_2C_2O_4$.
Let the oxidation number of carbon be $x$.
The sum of oxidation numbers of all atoms in a neutral molecule is $0$.
$2(+1) + 2(x) + 4(-2) = 0$
$2 + 2x - 8 = 0$
$2x - 6 = 0$
$2x = 6$
$x = +3$.

(C) Step $1$: Calculate the oxidation number of nitrogen in nitric acid $(HNO_3)$. Let the oxidation number of $N$ be $x$. Then, $1 + x + 3(-2) = 0$, which gives $x - 5 = 0$, so $x = +5$.
Step $2$: Calculate the oxidation number of nitrogen in nitrous oxide $(N_2O)$. Let the oxidation number of $N$ be $y$. Then, $2y + (-2) = 0$, which gives $2y = 2$, so $y = +1$.
Step $3$: Calculate the difference in oxidation numbers: $+5 - (+1) = +4$.

(B) Step $1$: Calculate the oxidation state of $Cl$ in $ClO_4^-$. Let the oxidation state be $x$. $x + 4(-2) = -1 \implies x - 8 = -1 \implies x = +7$.
Step $2$: Calculate the oxidation state of $Cl$ in $ClO_2^-$. Let the oxidation state be $y$. $y + 2(-2) = -1 \implies y - 4 = -1 \implies y = +3$.
Step $3$: Calculate the change in oxidation state per $Cl$ atom: $|7 - 3| = 4$.
Step $4$: For $n$ moles of $ClO_4^-$, the total number of electrons gained is $4 \times n = 4n$.

(B) Let the oxidation number of $Cl$ be $x$.
The oxidation number of $H$ is $+1$ and $O$ is $-2$.
For the neutral molecule $HOClO_2$, the sum of oxidation numbers is $0$.
$1 + x + 2(-2) = 0$
$1 + x - 4 = 0$
$x - 3 = 0$
$x = +5$.

(B) Step $1$: $A$ redox reaction involves a change in the oxidation state of elements.
Step $2$: In option $(A)$, $Zn$ changes from $0$ to $+2$ and $H$ changes from $+1$ to $0$.
Step $3$: In option $(B)$, $Al(OH)_3 + 3HCl \rightarrow AlCl_3 + 3H_2O$ is a neutralization reaction where oxidation states remain constant $(Al=+3, O=-2, H=+1, Cl=-1)$.
Step $4$: In option $(C)$, $Mg$ changes from $0$ to $+2$ and $O$ changes from $0$ to $-2$.
Step $5$: In option $(D)$, $Fe$ changes from $0$ to $+2$ and $Cu$ changes from $+2$ to $0$.
Step $6$: Therefore, option $(B)$ is not a redox reaction.

(A) $1$. In the reaction, $Sn^{2+}$ loses $2$ electrons to form $Sn^{4+}$, which is an increase in oxidation state (from $+2$ to $+4$), hence $Sn^{2+}$ is oxidized.
$2$. $Fe^{3+}$ gains electrons to form $Fe^{2+}$, which is a decrease in oxidation state (from $+3$ to $+2$), hence $Fe^{3+}$ is reduced.
$3$. Since both oxidation and reduction occur, it is a redox reaction.
$4$. Therefore, the correct statement is that $Sn^{2+}$ is undergoing oxidation.

(B) Oxidation is defined as:
$1$. Addition of oxygen or an electronegative element.
$2$. Removal of hydrogen or an electropositive element.
$3$. Loss of electrons.
Therefore, the addition of an electronegative element represents oxidation.

(D) Step $1$: $A$ redox reaction involves a change in the oxidation state of the participating elements.
Step $2$: In options $A$, $B$, and $C$, the oxidation states of the elements change, indicating electron transfer.
Step $3$: In option $D$, $HCl + H_2O \rightarrow H_3O^+ + Cl^-$, the oxidation state of $H$ remains $+1$ and $Cl$ remains $-1$ on both sides. This is an acid-base (proton transfer) reaction, not a redox reaction.

(B) Step $1$: Assign oxidation states to all atoms. In $H_3AsO_3$, $As$ is $+3$. In $BrO_3^-$, $Br$ is $+5$. In $Br^-$, $Br$ is $-1$. In $H_3AsO_4$, $As$ is $+5$.
Step $2$: Identify the oxidising agent. The oxidising agent is the species that gets reduced. Here, $Br$ in $BrO_3^-$ changes from $+5$ to $-1$, so $BrO_3^-$ is the oxidising agent.
Step $3$: The change in oxidation state of the oxidising agent is from $+5$ to $-1$.

(B) Using the ideal gas equation: $PV = nRT$
Rearranging for $n$: $n = \frac{PV}{RT}$
Given: $P = 1.104 \times 10^5 \text{ Nm}^{-2}$, $V = 4 \times 10^{-3} \text{ m}^3$, and $RT = 2208 \text{ J mol}^{-1}$.
Substituting the values: $n = \frac{(1.104 \times 10^5) \times (4 \times 10^{-3})}{2208}$
$n = \frac{110400 \times 0.004}{2208}$
$n = \frac{441.6}{2208} = 0.2 \text{ mole}$.

(D) According to Charles's Law, for a fixed mass of gas at constant pressure, $V \propto T$ (where $T$ is in Kelvin).
Step $1$: Convert temperatures to Kelvin: $T_1 = 10 + 273 = 283 \text{ K}$ and $T_2 = 20 + 273 = 293 \text{ K}$.
Step $2$: Use the formula $\frac{V_1}{T_1} = \frac{V_2}{T_2}$.
Step $3$: Substitute the values: $\frac{283 \text{ mL}}{283 \text{ K}} = \frac{V_2}{293 \text{ K}}$.
Step $4$: Solve for $V_2$: $V_2 = \frac{283 \times 293}{283} = 293 \text{ mL}$.

(A) The relationship between standard Gibbs energy change $(\Delta G^\circ)$ and equilibrium constant $(K)$ is given by: $\Delta G^\circ = -RT \ln K$.
Given: $R = 8.314 \text{ J K}^{-1} \text{mol}^{-1}$, $T = 298 \text{ K}$, $K = 100$.
$\Delta G^\circ = -8.314 \times 298 \times \ln(100)$.
Since $\ln(100) = 2.303 \times \log_{10}(100) = 2.303 \times 2 = 4.606$.
$\Delta G^\circ = -8.314 \times 298 \times 4.606 \text{ J/mol}$.
$\Delta G^\circ = -11411.4 \text{ J/mol} = -11.411 \text{ kJ/mol}$.

(B) The spontaneity of a reaction is determined by the Gibbs free energy change equation: $\Delta G = \Delta H - T\Delta S$.
For a reaction to be spontaneous, $\Delta G$ must be negative $(\Delta G < 0)$.
If $\Delta H < 0$ (exothermic) and $\Delta S > 0$ (increase in disorder), then $\Delta G = (\text{negative}) - T(\text{positive})$.
Since $T$ is always positive in Kelvin, $\Delta G$ will be negative at all temperatures.
Therefore, the correct condition is $\Delta H < 0$ and $\Delta S > 0$.

(A) The relationship between Gibbs free energy, enthalpy, and entropy is given by the equation: $\Delta G^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ}$.
Given $\Delta G^{\circ} = 0$, the equation becomes: $0 = \Delta H^{\circ} - T \Delta S^{\circ}$, which implies $T = \frac{\Delta H^{\circ}}{\Delta S^{\circ}}$.
Convert $\Delta H^{\circ}$ to $\text{J}$: $\Delta H^{\circ} = 40 \text{ kJ} = 40,000 \text{ J}$.
Substitute the values: $T = \frac{40,000 \text{ J}}{80 \text{ JK}^{-1}}$.
Calculate the result: $T = 500 \text{ K}$.

(D) The relationship between the standard Gibbs free energy change $\Delta G^{\circ}$ and the equilibrium constant $K_c$ is given by the equation:
$\Delta G^{\circ} = -RT \ln K_c$
Multiplying both sides by $-1$, we get:
$-\Delta G^{\circ} = RT \ln K_c$
Therefore, option $D$ is correct.

(A) The formula for standard Gibbs energy change is $\Delta G^\circ = -RT \ln K_p$.
Given: $R = 8.314 \text{ J K}^{-1} \text{mol}^{-1}$, $T = 298 \text{ K}$, $K_p = 3.5 \times 10^{17}$.
Using the relation $\ln x = 2.303 \log_{10} x$:
$\Delta G^\circ = -2.303 \times R \times T \times \log_{10} K_p$.
$\Delta G^\circ = -2.303 \times 8.314 \times 298 \times \log_{10}(3.5 \times 10^{17})$.
$\log_{10}(3.5 \times 10^{17}) = \log_{10}(3.5) + 17 \approx 0.544 + 17 = 17.544$.
$\Delta G^\circ = -2.303 \times 8.314 \times 298 \times 17.544 \approx -100135 \text{ J mol}^{-1}$.
Converting to kJ: $\Delta G^\circ \approx -100.1 \text{ kJ mol}^{-1}$.

(A) For a reaction to be spontaneous, the Gibbs free energy change $\Delta G$ must be negative, i.e., $\Delta G < 0$.
Given: $\Delta H = -50 \text{ kJ} = -50000 \text{ J}$ and $\Delta S = -100 \text{ J/K}$.
The formula is $\Delta G = \Delta H - T\Delta S$.
For spontaneity: $\Delta H - T\Delta S < 0$.
$-50000 - T(-100) < 0$.
$-50000 + 100T < 0$.
$100T < 50000$.
$T < 500 \text{ K}$.

(D) For a reaction to be at equilibrium, the Gibbs free energy change $\Delta G^0 = 0$.
The relationship is given by $\Delta G^0 = \Delta H^0 - T\Delta S^0$.
Setting $\Delta G^0 = 0$, we get $T = \frac{\Delta H^0}{\Delta S^0}$.
Convert $\Delta H^0$ to $\text{J}$: $\Delta H^0 = -345 \times 10^3 \text{ J}$.
Substitute the values: $T = \frac{-345 \times 10^3 \text{ J}}{-123 \text{ JK}^{-1}}$.
$T \approx 2804.87 \text{ K} \approx 2805 \text{ K}$.

(A) Step $1$: Identify bonds broken in reactants.
Bonds broken: $3 \times (C-H)$, $1 \times (C-O)$, $1 \times (O-H)$, $1 \times (H-Cl)$.
Energy required = $(3 \times 414) + 335 + 463 + 431 = 1242 + 335 + 463 + 431 = 2471 \text{ kJ/mol}$.
Step $2$: Identify bonds formed in products.
Bonds formed: $3 \times (C-H)$, $1 \times (C-Cl)$, $2 \times (H-O)$.
Energy released = $(3 \times 414) + 326 + (2 \times 463) = 1242 + 326 + 926 = 2494 \text{ kJ/mol}$.
Step $3$: Calculate enthalpy change $(\Delta H)$.
$\Delta H = \text{Energy of bonds broken} - \text{Energy of bonds formed}$.
$\Delta H = 2471 - 2494 = -23 \text{ kJ/mol}$.

(A) The chemical equation for the formation of $1 \text{ mole}$ of $HCl$ is: $\frac{1}{2} H_2(g) + \frac{1}{2} Cl_2(g) \rightarrow HCl(g)$.
Enthalpy of formation $(\Delta_f H^\circ)$ is calculated using bond enthalpies: $\Delta_f H^\circ = \sum \text{Bond Enthalpies of Reactants} - \sum \text{Bond Enthalpies of Products}$.
$\Delta_f H^\circ = [\frac{1}{2} \Delta H_{H-H} + \frac{1}{2} \Delta H_{Cl-Cl}] - [\Delta H_{H-Cl}]$.
Substituting the given values: $\Delta_f H^\circ = [\frac{1}{2}(434) + \frac{1}{2}(242)] - 431$.
$\Delta_f H^\circ = [217 + 121] - 431$.
$\Delta_f H^\circ = 338 - 431 = -93 \text{ kJ mol}^{-1}$.

(A) The enthalpy change of a reaction $(\Delta_r H^\circ)$ is calculated using bond energies as follows:
$\Delta_r H^\circ = \sum \text{Bond energies of reactants} - \sum \text{Bond energies of products}$
Step $1$: Calculate the energy required to break the reactant bonds:
Energy $= (1 \times \text{Bond energy of } H-H) + (1 \times \text{Bond energy of } Cl-Cl)$
Energy $= 436 + 242 = 678 \text{ kJ mol}^{-1}$
Step $2$: Calculate the energy released during the formation of product bonds:
Energy $= 2 \times \text{Bond energy of } H-Cl$
Energy $= 2 \times 431 = 862 \text{ kJ mol}^{-1}$
Step $3$: Calculate the enthalpy change:
$\Delta_r H^\circ = 678 - 862 = -184 \text{ kJ mol}^{-1}$

(A) The hydrogenation reaction of cyclohexene is: $C_6H_{10}(l) + H_2(g) \rightarrow C_6H_{12}(l)$.
Enthalpy of reaction $(\Delta H_{hydro})$ is calculated using enthalpies of combustion: $\Delta H_{hydro} = \sum \Delta H_c(\text{reactants}) - \sum \Delta H_c(\text{products})$.
$\Delta H_{hydro} = [\Delta H_c(C_6H_{10}) + \Delta H_c(H_2)] - [\Delta H_c(C_6H_{12})]$.
Substituting the given values: $\Delta H_{hydro} = [-3800 + (-241)] - [-3920]$.
$\Delta H_{hydro} = -4041 + 3920 = -121 \text{ kJ mol}^{-1}$.

(B) The relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta U)$ is given by: $\Delta H = \Delta U + \Delta n_g RT$.
Here, $\Delta n_g$ is the change in the number of moles of gaseous products and reactants.
For the reaction: $C_2H_5OH(l) + 3O_2(g) \rightarrow 2CO_2(g) + 3H_2O(l)$.
$\Delta n_g = (n_{g, \text{products}}) - (n_{g, \text{reactants}}) = 2 - 3 = -1$.
Given: $\Delta H = -1366.5 \text{ kJ mol}^{-1}$, $T = 27 + 273 = 300 \text{ K}$, $R = 8.314 \times 10^{-3} \text{ kJ K}^{-1} \text{ mol}^{-1}$.
Substituting the values: $-1366.5 = \Delta U + (-1) \times (8.314 \times 10^{-3}) \times 300$.
$-1366.5 = \Delta U - 2.4942$.
$\Delta U = -1366.5 + 2.4942 = -1364.0058 \text{ kJ mol}^{-1} \approx -1364.0 \text{ kJ mol}^{-1}$.

(D) The relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta U)$ is given by the equation: $\Delta H = \Delta U + \Delta n_g RT$.
For the reaction $S(s) + O_2(g) \rightarrow SO_2(g)$, the change in the number of moles of gaseous species is $\Delta n_g = n_p(g) - n_r(g)$.
Here, $n_p(g) = 1$ (for $SO_2$) and $n_r(g) = 1$ (for $O_2$).
Therefore, $\Delta n_g = 1 - 1 = 0$.
Substituting this into the equation: $\Delta H = \Delta U + (0)RT = \Delta U$.
Thus, $\Delta H = \Delta U$.

(B) Step $1$: Identify the sign conventions for heat $(q)$ and work $(w)$.
Heat released by the system: $q = -4 \text{ kJ}$.
Work done on the system: $w = +12 \text{ kJ}$.
Step $2$: Calculate the change in internal energy $(\Delta U)$ using the first law of thermodynamics: $\Delta U = q + w$.
$\Delta U = -4 \text{ kJ} + 12 \text{ kJ} = 8 \text{ kJ}$.
Step $3$: For a reaction at constant pressure, the heat released is equal to the change in enthalpy $(\Delta H)$.
$\Delta H = q_p = -4 \text{ kJ}$.
Thus, $\Delta H = -4 \text{ kJ}$ and $\Delta U = 8 \text{ kJ}$.

(D) $1$. In thermodynamics, a process where there is no exchange of heat $(q = 0)$ between the system and the surroundings is defined as an adiabatic process.
$2$. For an adiabatic process, the first law of thermodynamics simplifies to $\Delta U = w$, where $\Delta U$ is the change in internal energy and $w$ is the work done.
$3$. Therefore, the correct option is $D$.

(B) According to the first law of thermodynamics: $\Delta U = q + w$
Given:
Heat provided to the system, $q = +50 \text{ J}$
Work done on the system, $w = +10 \text{ J}$
Substituting the values:
$\Delta U = 50 \text{ J} + 10 \text{ J} = 60 \text{ J}$
Thus, the change in internal energy is $60 \text{ J}$.

(C) For an isothermal process of an ideal gas, the change in internal energy $\Delta U = 0$.
According to the first law of thermodynamics, $\Delta U = q + w$, so $q = -w$.
The work done on the system during compression against a constant external pressure is given by $w = -P_{\text{ext}} \Delta V$.
Here, $P_{\text{ext}} = 4 \text{ bar} = 4 \times 10^5 \text{ Pa}$, $V_1 = 25 \text{ dm}^3 = 25 \times 10^{-3} \text{ m}^3$, and $V_2 = 10 \text{ dm}^3 = 10 \times 10^{-3} \text{ m}^3$.
$\Delta V = V_2 - V_1 = (10 - 25) \times 10^{-3} \text{ m}^3 = -15 \times 10^{-3} \text{ m}^3$.
$w = -(4 \times 10^5 \text{ Pa}) \times (-15 \times 10^{-3} \text{ m}^3) = 6000 \text{ J} = 6 \text{ kJ}$.
Since $q = -w$, the heat released is $q = -6 \text{ kJ}$.
The quantity of heat released is $6 \text{ kJ}$.

(A) The first law of thermodynamics is given by $\Delta U = Q + W$.
For an isothermal process, the temperature remains constant, which implies that the internal energy change $\Delta U = 0$.
Substituting $\Delta U = 0$ into the first law equation: $0 = Q + W$.
Rearranging the terms, we get $W = -Q$.

(A) The relationship between $\Delta H$ and $\Delta U$ is given by: $\Delta H = \Delta U + \Delta n_g RT$
For the reaction $2C(s) + 3H_2(g) \rightarrow C_2H_6(g)$:
Calculate the change in the number of gaseous moles, $\Delta n_g = n_p(g) - n_r(g) = 1 - 3 = -2 \text{ mol}$.
Given $\Delta U = -80 \text{ kJ} = -80000 \text{ J}$, $T = 300 \text{ K}$, and $R = 8.314 \text{ J K}^{-1} \text{mol}^{-1}$.
$\Delta H = -80000 \text{ J} + (-2 \text{ mol} \times 8.314 \text{ J K}^{-1} \text{mol}^{-1} \times 300 \text{ K})$
$\Delta H = -80000 \text{ J} - 4988.4 \text{ J} = -84988.4 \text{ J} \approx -85.00 \text{ kJ}$.

(B) The first law of thermodynamics is given by $\Delta U = Q + W$.
For an adiabatic process, there is no exchange of heat between the system and the surroundings, so $Q = 0$.
Substituting $Q = 0$ into the first law equation, we get $\Delta U = W$.
Alternatively, if work is done by the system ($W$ is negative), then $\Delta U = -W$, which can be written as $-\Delta U = W$ or $\Delta U = -W_{on}$.
Looking at the options, $-\Delta U = -W$ simplifies to $\Delta U = W$, which is the correct mathematical representation for an adiabatic process where $W$ is the work done on the system.

(A) For an isothermal reversible compression of an ideal gas, the work done $W$ is given by the formula:
$W = -nRT \ln\left(\frac{P_1}{P_2}\right)$
Given:
$n = 1 \text{ mol}$
$R = 8.314 \text{ J K}^{-1} \text{mol}^{-1}$
$T = 300 \text{ K}$
$P_1 = x \text{ kPa}$
$P_2 = 2x \text{ kPa}$
Substituting the values:
$W = -(1 \text{ mol}) \times (8.314 \text{ J K}^{-1} \text{mol}^{-1}) \times (300 \text{ K}) \times \ln\left(\frac{x}{2x}\right)$
$W = -2494.2 \times \ln(0.5)$
$W = -2494.2 \times (-0.693)$
$W \approx 1729 \text{ J}$
Since the gas is compressed, work is done on the system, hence the value is positive.

(A) The formula for work done during expansion against constant external pressure is $W = -P_{\text{ext}} \Delta V$.
Given:
$P_{\text{ext}} = 2 \text{ atm}$
$V_1 = 5 \text{ dm}^3$, $V_2 = 15 \text{ dm}^3$
$\Delta V = V_2 - V_1 = 15 \text{ dm}^3 - 5 \text{ dm}^3 = 10 \text{ dm}^3 = 10 \text{ L}$.
Since $1 \text{ L atm} = 101.325 \text{ J}$,
$W = -2 \text{ atm} \times 10 \text{ L} = -20 \text{ L atm}$.
$W = -20 \times 101.32 \text{ J} = -2026.4 \text{ J}$.
Thus, the work done by the gas is $-2026.4 \text{ J}$.

(D) According to the first law of thermodynamics, $\Delta U = q + w$.
Given: Heat absorbed $q = +120 \text{ J}$.
External pressure $P_{ext} = 2 \times 10^5 \text{ Nm}^{-2}$.
Change in volume $\Delta V = 300 \times 10^{-6} \text{ m}^3$.
Work done by the system $w = -P_{ext} \times \Delta V$.
$w = -(2 \times 10^5 \text{ Nm}^{-2}) \times (300 \times 10^{-6} \text{ m}^3) = -60 \text{ J}$.
Change in internal energy $\Delta U = 120 \text{ J} + (-60 \text{ J}) = 60 \text{ J}$.

(B) $1$. In an $H_2 - O_2$ fuel cell, carbon rods are used as electrodes.
$2$. These carbon rods are impregnated with finely divided platinum or palladium catalyst to increase the rate of reaction.
$3$. $A$ hot concentrated aqueous solution of $KOH$ (potassium hydroxide) is used as the electrolyte in this fuel cell.
$4$. The electrolyte facilitates the movement of ions between the electrodes.

(D) In a lead storage battery acting as a galvanic cell (discharging), the positive electrode is $PbO_2$.
At the positive electrode, the reduction reaction is: $PbO_2(s) + SO_4^{2-}(aq) + 4H^+(aq) + 2e^- \rightarrow PbSO_4(s) + 2H_2O(l)$.
In $PbO_2$, the oxidation number of $Pb$ is $+4$.
In $PbSO_4$, the oxidation number of $Pb$ is $+2$.
Change in oxidation number = $+2 - (+4) = -2$.
Thus, the oxidation number decreases by $2$.

(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 occurs:
$PbSO_4(s) + 2e^- \rightarrow Pb(s) + SO^{2-}_{4(aq)}$
At the anode (positive electrode during recharging), the oxidation reaction occurs:
$PbSO_4(s) + 2H_2O(\ell) \rightarrow PbO_2(s) + SO^{2-}_{4(aq)} + 4H^+_{(aq)} + 2e^-$
Therefore, the reaction at the cathode is $PbSO_4(s) + 2e^- \rightarrow Pb(s) + SO^{2-}_{4(aq)}$.

(C) $1$. $A$ salt bridge is a $U$-shaped tube containing an electrolyte (like $KCl$ or $KNO_3$) in a gel.
$2$. It connects the two half-cells of an electrochemical cell.
$3$. Its primary functions are to complete the electrical circuit and maintain electrical neutrality in both half-cells by allowing the migration of ions.
$4$. Therefore, the correct function is to connect two half-cells.

(D) During the electrolysis of molten $NaCl$, the following reactions occur:
At anode (oxidation): $2Cl^- (l) \rightarrow Cl_2 (g) + 2e^-$. Pale green $Cl_2$ gas is released.
At cathode (reduction): $Na^+ (l) + e^- \rightarrow Na (l)$. Molten silvery-white sodium metal is deposited.
Since both statements $A$ and $B$ are correct, the overall process is the decomposition of $NaCl$ into $Na$ metal and $Cl_2$ gas, making option $D$ the most comprehensive correct choice.

(A) Step $1$: Calculate Faraday for each conversion.
$A$. $H_2O \rightarrow \frac{1}{2} O_2 + 2H^+ + 2e^-$. Thus, $2 \text{ moles of } e^-$ are required, which is $2F$.
$B$. $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$. Thus, $5 \text{ moles of } e^-$ are required, which is $5F$. (Note: The provided options suggest a match with $IV$ $(5F)$ for $B$).
$C$. $Ca^{2+} + 2e^- \rightarrow Ca$. For $1 \text{ mol of } Ca$, $2F$ is needed. For $1.5 \text{ mol}$, $1.5 \times 2 = 3F$. Thus, $C-I$.
$D$. $FeO \rightarrow \frac{1}{2} Fe_2O_3$. Oxidation state of $Fe$ changes from $+2$ to $+3$. Change is $1e^-$. Thus, $1F$. So, $D-III$.
Correct matching: $A-II, B-IV, C-I, D-III$.

(C) The Nernst equation for the electrode reaction $Cu^{2+} + 2e^- \rightarrow Cu(s)$ is:
$E_{Cu^{2+}/Cu} = E^{\circ}_{Cu^{2+}/Cu} - \frac{0.0591}{n} \log \frac{1}{[Cu^{2+}]}$
Here, $E^{\circ}_{Cu^{2+}/Cu} = 0.34 \text{ V}$, $n = 2$, and $[Cu^{2+}] = 0.1 \text{ M}$.
$E_{Cu^{2+}/Cu} = 0.34 - \frac{0.0591}{2} \log \frac{1}{0.1}$
$E_{Cu^{2+}/Cu} = 0.34 - 0.02955 \times \log(10)$
Since $\log(10) = 1$, we get:
$E_{Cu^{2+}/Cu} = 0.34 - 0.02955 = 0.31045 \text{ V} \approx 0.31 \text{ V}$.

(B) In an electrochemical cell, the electrode where reduction takes place is defined as the cathode. Electrons flow towards the cathode, and it is typically the positive electrode in a galvanic cell. Therefore, reduction occurs at the cathode.

(A) reaction is spontaneous if the standard cell potential $E^0_{cell} > 0$. The formula is $E^0_{cell} = E^0_{cathode} - E^0_{anode}$.
For $(A)$: $Ca$ is oxidized (anode), $Cd^{2+}$ is reduced (cathode). $E^0_{cell} = (-0.403) - (-2.866) = +2.463 \text{ V}$. Since $E^0_{cell} > 0$, the reaction is spontaneous.
For $(B)$: $Br^-$ is oxidized (anode), $Sn^{2+}$ is reduced (cathode). $E^0_{cell} = (-0.136) - (1.08) = -1.216 \text{ V}$. Non-spontaneous.
For $(C)$: $Ag$ is oxidized (anode), $Ni^{2+}$ is reduced (cathode). $E^0_{cell} = (-0.257) - (0.799) = -1.056 \text{ V}$. Non-spontaneous.
For $(D)$: $Au$ is oxidized (anode), $Zn^{2+}$ is reduced (cathode). $E^0_{cell} = (-0.763) - (1.68) = -2.443 \text{ V}$. Non-spontaneous.
Therefore, option $(A)$ is correct.

(C) $1$. Anode is the electrode where oxidation occurs (loss of electrons).
$2$. In option $(A)$, $F_2$ gains electrons (reduction).
$3$. In option $(B)$, $H^+$ and $O_2$ gain electrons (reduction).
$4$. In option $(C)$, $Fe^{2+}$ loses an electron to form $Fe^{3+}$, which is an oxidation process.
$5$. In option $(D)$, $Cu^{2+}$ gains electrons (reduction).
$6$. Therefore, only reaction $(C)$ represents oxidation and can occur at the anode.

(B) The cell reaction is: $2Al(s) + 3Pb^{2+}(aq) \rightarrow 2Al^{3+}(aq) + 3Pb(s)$.
The standard cell potential is given by: $E^0_{cell} = E^0_{cathode} - E^0_{anode}$.
Here, the cathode is $Pb^{2+}/Pb$ and the anode is $Al^{3+}/Al$.
So, $E^0_{cell} = E^0_{Pb^{2+}/Pb} - E^0_{Al^{3+}/Al}$.
Given $E^0_{cell} = 1.5 \text{ V}$ and $E^0_{Pb^{2+}/Pb} = -0.14 \text{ V}$.
Substituting the values: $1.5 \text{ V} = -0.14 \text{ V} - E^0_{Al^{3+}/Al}$.
$E^0_{Al^{3+}/Al} = -0.14 \text{ V} - 1.5 \text{ V} = -1.64 \text{ V}$.

(D) According to Faraday's second law of electrolysis, the mass of a substance deposited is proportional to its equivalent mass.
Equivalent mass $E = \frac{\text{Atomic mass}}{\text{Valency factor (n-factor)}}$.
For $FeCl_2$, iron is in $+2$ oxidation state, so $n_A = 2$. $E_A = \frac{M}{2}$.
For $FeCl_3$, iron is in $+3$ oxidation state, so $n_B = 3$. $E_B = \frac{M}{3}$.
The ratio of masses deposited is $\frac{m_A}{m_B} = \frac{E_A}{E_B} = \frac{M/2}{M/3} = \frac{3}{2}$.
Thus, the ratio is $3 : 2$.

(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 anions $(Cl^-)$.
$3$. Oxidation occurs at the anode, where chloride ions lose electrons to form chlorine gas.
$4$. The half-reaction at the anode is: $2Cl^-_{(l)} \rightarrow Cl_{2(g)} + 2e^-$.

(B) $1$. The standard electrode potential $(E^\circ)$ is defined for the reduction half-reaction.
$2$. The notation $Cu^{+2}/Cu$ represents the reduction of $Cu^{+2}$ ions to metallic $Cu$.
$3$. The corresponding reduction half-reaction is $Cu^{+2} + 2e^- \rightarrow Cu$.
$4$. Therefore, the correct reaction is option $B$.

(D) Step $1$: Calculate the moles of $O_2$ gas. At $STP$, $22400 \text{ mL}$ corresponds to $1 \text{ mole}$. Thus, moles of $O_2 = \frac{5600 \text{ mL}}{22400 \text{ mL/mol}} = 0.25 \text{ mol}$.
Step $2$: Determine the total charge required. The reaction for $O_2$ evolution is $2H_2O \rightarrow O_2 + 4H^+ + 4e^-$. This shows $1 \text{ mole}$ of $O_2$ requires $4 \text{ moles}$ of electrons. Therefore, $0.25 \text{ mol}$ of $O_2$ requires $0.25 \times 4 = 1 \text{ mole}$ of electrons.
Step $3$: Calculate the mass of silver deposited. The reaction for silver deposition is $Ag^+ + e^- \rightarrow Ag$. This shows $1 \text{ mole}$ of electrons deposits $1 \text{ mole}$ of $Ag$. Mass of $Ag = 1 \text{ mol} \times 108 \text{ g/mol} = 108 \text{ g}$.

(C) The cell reaction is: $Cu(s) + 2Ag^+(aq) \rightarrow Cu^{2+}(aq) + 2Ag(s)$.
The standard cell potential is given by $E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$.
Since the concentrations are $1M$, the cell potential $E_{cell}$ is equal to the standard cell potential $E^{\circ}_{cell} = 0.463 \text{ V}$.
Here, $Ag$ is the cathode and $Cu$ is the anode.
$E^{\circ}_{cell} = E^{\circ}_{Ag^+/Ag} - E^{\circ}_{Cu^{2+}/Cu}$.
$0.463 \text{ V} = E^{\circ}_{Ag^+/Ag} - 0.337 \text{ V}$.
$E^{\circ}_{Ag^+/Ag} = 0.463 \text{ V} + 0.337 \text{ V} = 0.800 \text{ V}$.

(A) reaction is non-spontaneous if the standard cell potential $E^0_{cell} < 0$.
$E^0_{cell} = E^0_{cathode} - E^0_{anode}$.
For option $(A)$: $2Ag(s) + Fe^{+2}_{(aq)} \rightarrow 2Ag^{+1}_{(aq)} + Fe(s)$.
$E^0_{cathode} = E^0_{(Fe^{+2}|Fe)} = -0.44 \text{ V}$.
$E^0_{anode} = E^0_{(Ag^{+1}|Ag)} = +0.79 \text{ V}$.
$E^0_{cell} = -0.44 - 0.79 = -1.23 \text{ V}$.
Since $E^0_{cell} < 0$, the reaction is non-spontaneous.

(B) $1$. The standard reduction potential of the standard hydrogen electrode $(SHE)$ is $E^0_{(H^+|H_2)} = 0.00 \text{ V}$.
$2$. The standard reduction potential of the cadmium electrode is $E^0_{(Cd^{2+}|Cd)} = -0.403 \text{ V}$.
$3$. Since $E^0_{(H^+|H_2)} > E^0_{(Cd^{2+}|Cd)}$, the hydrogen electrode acts as the cathode (reduction) and the cadmium electrode acts as the anode (oxidation).
$4$. Anode reaction: $Cd(s) \rightarrow Cd^{2+}_{(aq)} + 2e^-$.
$5$. Cathode reaction: $2H^+_{(aq)} + 2e^- \rightarrow H_2(g)$.
$6$. Adding these two half-reactions gives the net cell reaction: $Cd(s) + 2H^+_{(aq)} \rightarrow Cd^{2+}_{(aq)} + H_2(g)$.

(B) The relationship between $E^0_{cell}$ and the equilibrium constant $K_c$ is given by the Nernst equation at $298 \text{ K}$:
$E^0_{cell} = \frac{0.0591 \text{ V}}{n} \log K_c$
Here, $n = 2$ (number of electrons transferred).
$0.0296 = \frac{0.0591}{2} \log K_c$
$0.0296 = 0.02955 \log K_c$
Since $0.0296 \approx 0.02955$, we get $\log K_c \approx 1$.
$K_c = 10^1 = 10$.

(C) The standard $emf$ of a cell is given by $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$.
To maximize $E^\circ_{\text{cell}}$, we must choose the cathode with the highest reduction potential and the anode with the lowest reduction potential.
Given potentials: $E^\circ_A = +2.5 \text{ V}$, $E^\circ_B = +3.0 \text{ V}$, $E^\circ_C = -2.0 \text{ V}$, $E^\circ_D = -1.5 \text{ V}$.
Highest potential (cathode) is $B$ $(+3.0 \text{ V})$.
Lowest potential (anode) is $C$ $(-2.0 \text{ V})$.
$E^\circ_{\text{cell}} = E^\circ_B - E^\circ_C = 3.0 - (-2.0) = 5.0 \text{ V}$.
Thus, $C$ is the anode and $B$ is the cathode.

(D) For a reaction to be spontaneous, the standard cell potential $E^\circ_{\text{cell}}$ must be positive.
$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = E^\circ_{Y^{+2}/Y} - E^\circ_{X^{+2}/X} > 0$.
This implies $E^\circ_{Y^{+2}/Y} > E^\circ_{X^{+2}/X}$.
Given potentials: $E^\circ_{Zn^{+2}/Zn} = -0.76 \text{ V}$, $E^\circ_{Fe^{+2}/Fe} = -0.44 \text{ V}$, $E^\circ_{Ni^{+2}/Ni} = -0.23 \text{ V}$.
Checking option $(D)$: $X = Zn$ $(E^\circ = -0.76 \text{ V})$ and $Y = Ni$ $(E^\circ = -0.23 \text{ V})$.
$E^\circ_{\text{cell}} = (-0.23) - (-0.76) = +0.53 \text{ V}$.
Since $E^\circ_{\text{cell}} > 0$, the reaction is spontaneous.

(B) $1$. In a galvanic cell, the electrode with the higher reduction potential acts as the cathode (positive electrode).
$2$. Given $E^0_{(Cu^{2+}|Cu)} = +0.34 \text{ V}$ and $E^0_{(H^+|H_2)} = 0.00 \text{ V}$.
$3$. Since $+0.34 \text{ V} > 0.00 \text{ V}$, the copper electrode is the cathode.
$4$. Reduction occurs at the cathode: $Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)$.

(D) $1$. In an aqueous solution of $NaCl$, the ions present are $Na^+$, $Cl^-$, $H^+$, and $OH^-$.
$2$. At the cathode, the reduction reaction competes between $Na^+ + e^- \rightarrow Na(s)$ $(E^o = -2.71 \text{ V})$ and $2H_2O(l) + 2e^- \rightarrow H_{2(g)} + 2OH^-(aq)$ $(E^o = -0.83 \text{ V})$.
$3$. Since the reduction potential of water is higher than that of $Na^+$, water is reduced preferentially.
$4$. Therefore, $H_{2(g)}$ is produced at the cathode.

(A) Step $1$: Calculate the total charge $Q$ passed using the formula $Q = I \times t$.
Given: $I = 2 \text{ A}$, $t = 10 \text{ minutes} = 10 \times 60 \text{ s} = 600 \text{ s}$.
$Q = 2 \text{ A} \times 600 \text{ s} = 1200 \text{ C}$.
Step $2$: Calculate the number of moles of electrons $(n)$ using the relation $n = \frac{Q}{F}$, where $F$ is Faraday's constant $(96500 \text{ C/mol})$.
$n = \frac{1200 \text{ C}}{96500 \text{ C/mol}} \approx 0.012435 \text{ mol}$.
Step $3$: Express in scientific notation: $n \approx 1.243 \times 10^{-2} \text{ mol}$.

(C) The chemical reaction for the reduction of $Mg^{2+}$ to $Mg$ is: $Mg^{2+} + 2e^- \rightarrow Mg(s)$.
From the stoichiometry, $1 \text{ mole}$ of $Mg$ $(24 \text{ g})$ requires $2 \text{ moles}$ of electrons, which is equal to $2 \text{ F}$ of electricity.
Number of moles of $Mg$ produced = $\frac{\text{mass}}{\text{molar mass}} = \frac{8 \text{ g}}{24 \text{ g mol}^{-1}} = \frac{1}{3} \text{ mol}$.
Electricity required = $(\text{moles of } Mg) \times 2 \text{ F/mol} = \frac{1}{3} \times 2 \text{ F} = 0.666 \text{ F}$.

(C) The reduction half-reaction is: $Sn^{2+}(aq) + 2e^- \rightarrow Sn(s)$.
Using the Nernst equation: $E_{Sn^{2+}/Sn} = E^0_{Sn^{2+}/Sn} - \frac{0.0591}{n} \log \frac{1}{[Sn^{2+}]}$.
Here, $n = 2$, $E^0 = -0.136 \text{ V}$, and $[Sn^{2+}] = 0.01 \text{ M} = 10^{-2} \text{ M}$.
$E = -0.136 - \frac{0.0591}{2} \log \frac{1}{10^{-2}}$.
$E = -0.136 - 0.02955 \times \log(10^2)$.
$E = -0.136 - 0.02955 \times 2$.
$E = -0.136 - 0.0591 = -0.1951 \text{ V}$.
Rounding to three decimal places, $E = -0.195 \text{ V}$.

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

(B) The given compound is $CH_3(CH_2)_2COOH$, which is $CH_3CH_2CH_2COOH$ (Butanoic acid).
Homologues differ by a $-CH_2-$ unit.
The first lower homologue is $CH_3CH_2COOH$ (Propanoic acid).
The second lower homologue is $CH_3COOH$ (Ethanoic acid).
Therefore, the correct option is $B$.

(C) $1$. Formic acid is the first member of the carboxylic acid homologous series with the formula $HCOOH$ (Methanoic acid).
$2$. The first higher homologue is obtained by adding a $-CH_2-$ group, which is $CH_3COOH$ (Ethanoic acid).
$3$. The second higher homologue is obtained by adding another $-CH_2-$ group to the first higher homologue, which is $CH_3CH_2COOH$ (Propanoic acid).
$4$. Therefore, the second higher homologue of formic acid is Propanoic acid.

(D) An aromatic aldehyde is a compound in which the $-CHO$ group is directly attached to an aromatic ring (benzene ring).
$(A)$ Phenylacetaldehyde: The $-CHO$ group is attached to an alkyl chain, not directly to the benzene ring.
$(B)$ Cinnamaldehyde: The $-CHO$ group is attached to a conjugated side chain, not directly to the benzene ring.
$(C)$ Cyclohexanecarbaldehyde: The ring is non-aromatic (cyclohexane).
$(D)$ Benzaldehyde: The $-CHO$ group is directly attached to the benzene ring, making it an aromatic aldehyde.
Therefore, the correct option is $(D)$.

(C) Frenkel defect is observed in ionic crystals where there is a large difference in the size of the ions. Specifically, it occurs when the smaller ion (usually the cation) leaves its lattice site and occupies an interstitial site.
$AgCl$ exhibits Frenkel defect because the $Ag^+$ ion is significantly smaller than the $Cl^-$ ion, allowing it to move into an interstitial position.
$NaCl$, $KCl$, and $CsCl$ typically exhibit Schottky defect due to the similar sizes of their constituent ions.

(D) Step $1$: Identify the number of atoms per unit cell for $bcc$. For $bcc$, $Z = 2$.
Step $2$: Use the density formula: $\rho = \frac{Z \times M}{N_A \times a^3}$, which can be rearranged as $M = \frac{\rho \times a^3 \times N_A}{Z}$.
Step $3$: Calculate the molar mass $M$: $M = \frac{3.25 \times 10^{-22} \times 6.022 \times 10^{23}}{2} = \frac{195.715}{2} \approx 97.86 \text{ g/mol}$.
Step $4$: Calculate the number of moles $n$: $n = \frac{\text{mass}}{M} = \frac{1.625}{97.86} \approx 0.0166 \text{ mol}$.
Step $5$: Calculate the number of atoms: $\text{Atoms} = n \times N_A = 0.0166 \times 6.022 \times 10^{23} \approx 1.0 \times 10^{22}$ atoms.

(C) In a face-centered cubic $(fcc)$ unit cell, the number of atoms per unit cell is $Z = 4$.
The number of tetrahedral voids is equal to $2 \times Z$.
Therefore, the number of tetrahedral voids $= 2 \times 4 = 8$.
The number of tetrahedral voids per sphere (atom) $= \frac{8}{4} = 2$.

(B) $1$. For a $bcc$ structure, the number of atoms per unit cell, $Z = 2$.
$2$. The mass of one unit cell is given by $m = \rho \times V = \rho \times a^3$.
$3$. Given $\rho \times a^3 = 1.8 \times 10^{-22} \text{ g}$, so the mass of one unit cell is $1.8 \times 10^{-22} \text{ g}$.
$4$. The number of unit cells in $90 \text{ g}$ of the metal is $N_{uc} = \frac{\text{Total mass}}{\text{Mass of one unit cell}} = \frac{90 \text{ g}}{1.8 \times 10^{-22} \text{ g}} = 50 \times 10^{22} = 5.0 \times 10^{23}$.
$5$. Total number of atoms = $Z \times N_{uc} = 2 \times 5.0 \times 10^{23} = 1.0 \times 10^{24}$ atoms.

(B) $1$. In an $hcp$ structure, if the number of atoms is $N$, then the number of tetrahedral voids is $2N$.
$2$. The number of atoms in $0.8 \text{ mol}$ is $0.8 \times N_A$, where $N_A = 6.022 \times 10^{23} \text{ mol}^{-1}$.
$3$. Number of tetrahedral voids $= 2 \times (0.8 \times N_A) = 1.6 \times N_A$.
$4$. Calculation: $1.6 \times 6.022 \times 10^{23} = 9.6352 \times 10^{23}$.
$5$. Thus, the correct option is $B$.

(B) $1$. $AgCl$ crystallizes in a rock salt $(NaCl)$ type structure.
$2$. In this structure, each $Ag^+$ ion is surrounded by $6$ $Cl^-$ ions, and each $Cl^-$ ion is surrounded by $6$ $Ag^+$ ions.
$3$. Therefore, the coordination number of $Ag^+$ in $AgCl$ is $6$.

(B) $1$. For a face-centered cubic $(FCC)$ unit cell, the number of atoms per unit cell, $Z = 4$.
$2$. The formula for density $(d)$ is $d = \frac{Z \times M}{a^3 \times N_A}$, where $a^3$ is the volume of the unit cell.
$3$. Given: $Z = 4$, $M = 108 \text{ g/mol}$, $V = a^3 = 6.84 \times 10^{-23} \text{ cm}^3$, $N_A = 6.022 \times 10^{23} \text{ mol}^{-1}$.
$4$. Substituting the values: $d = \frac{4 \times 108}{6.84 \times 10^{-23} \times 6.022 \times 10^{23}}$.
$5$. $d = \frac{432}{6.84 \times 6.022} = \frac{432}{41.19} \approx 10.49 \text{ g cm}^{-3}$.

(A) In a $bcc$ (body-centered cubic) unit cell, there are $8$ corner particles.
Each corner particle is shared by $8$ adjacent unit cells.
Therefore, the contribution of each corner particle to a single unit cell is $\frac{1}{8}$.
Total contribution of all corner particles = $8 \times \frac{1}{8} = 1 \text{ particle}$.

(C) Step $1$: Convert the edge length $a$ from $\text{pm}$ to $\text{cm}$.
$a = 500 \text{ pm} = 500 \times 10^{-10} \text{ cm} = 5 \times 10^{-8} \text{ cm}$.
Step $2$: Calculate the volume of one unit cell $(V_{uc} = a^3)$.
$V_{uc} = (5 \times 10^{-8} \text{ cm})^3 = 125 \times 10^{-24} \text{ cm}^3 = 1.25 \times 10^{-22} \text{ cm}^3$.
Step $3$: Calculate the number of unit cells in $1 \text{ cm}^3$.
$\text{Number of unit cells} = \frac{\text{Total volume}}{\text{Volume of one unit cell}} = \frac{1 \text{ cm}^3}{1.25 \times 10^{-22} \text{ cm}^3} = 0.8 \times 10^{22} = 8.0 \times 10^{21}$.

(D) The volume $V$ of a spherical particle is given by the formula $V = \frac{4}{3} \pi r^3$.
Given radius $r = 190 \text{ pm} = 190 \times 10^{-10} \text{ cm} = 1.9 \times 10^{-8} \text{ cm}$.
Substituting the value of $r$ into the formula:
$V = \frac{4}{3} \times 3.14159 \times (1.9 \times 10^{-8} \text{ cm})^3$.
$V = \frac{4}{3} \times 3.14159 \times 6.859 \times 10^{-24} \text{ cm}^3$.
$V = 4.18879 \times 6.859 \times 10^{-24} \text{ cm}^3$.
$V \approx 28.73 \times 10^{-24} \text{ cm}^3 = 2.873 \times 10^{-23} \text{ cm}^3$.

(D) $1$. The packing efficiency of a $bcc$ unit cell is $68\%$.
$2$. The volume occupied by atoms = $0.68 \times \text{Total volume of unit cell}$.
$3$. Volume occupied = $0.68 \times 8.0 \times 10^{-23} \text{ cm}^3 = 5.44 \times 10^{-23} \text{ cm}^3$.
$4$. Void volume = $\text{Total volume} - \text{Volume occupied}$.
$5$. Void volume = $8.0 \times 10^{-23} - 5.44 \times 10^{-23} = 2.56 \times 10^{-23} \text{ cm}^3$.

(C) For a $bcc$ crystal structure, the number of atoms per unit cell, $Z = 2$.
Density formula is given by $\rho = \frac{Z \times M}{N_A \times a^3}$, where $M$ is the molar mass and $N_A$ is Avogadro's number.
Rearranging for the mass of one unit cell: $m_{cell} = \rho \times a^3 = \frac{Z \times M}{N_A}$.
Given $\rho \times a^3 = 6.6 \times 10^{-22} \text{ g}$, this represents the mass of one unit cell.
The number of unit cells in $1 \text{ g}$ of metal is $n = \frac{1 \text{ g}}{6.6 \times 10^{-22} \text{ g/unit cell}} \approx 1.515 \times 10^{21} \text{ unit cells}$.
Since each $bcc$ unit cell contains $2$ atoms, the total number of atoms $= 2 \times 1.515 \times 10^{21} = 3.03 \times 10^{21} \text{ atoms}$.

(A) Step $1$: Identify the number of atoms per unit cell $(Z)$ for a simple cubic structure. For a simple cubic unit cell, $Z = 1$.
Step $2$: Use the density formula: $\rho = \frac{Z \times M}{N_A \times V}$, where $\rho$ is density, $M$ is molar mass, $N_A$ is Avogadro's number, and $V$ is the volume of the unit cell.
Step $3$: Rearrange to find the mass of one unit cell: $\rho \times V = \frac{Z \times M}{N_A}$. Given $\rho \times V = 66.5 \times 10^{-24} \text{ g}$, this represents the mass of one unit cell.
Step $4$: Calculate the number of unit cells in $1 \text{ g}$ of metal: $\text{Number of unit cells} = \frac{\text{Total mass}}{\text{Mass of one unit cell}} = \frac{1 \text{ g}}{66.5 \times 10^{-24} \text{ g}} \approx 1.5037 \times 10^{22}$.
Step $5$: Since each unit cell contains $1$ atom $(Z=1)$, the total number of atoms is $1.5037 \times 10^{22} \times 1 = 1.504 \times 10^{22}$.

(C) $1$. The packing efficiency of an $fcc$ unit cell is $74\%$. This means $74\%$ of the total volume is occupied by atoms.
$2$. The void volume is the remaining volume, which is $100\% - 74\% = 26\%$ of the total volume.
$3$. Given total volume $V_{total} = 6.4 \times 10^{-23} \text{ cm}^3$.
$4$. Void volume $= 0.26 \times V_{total} = 0.26 \times 6.4 \times 10^{-23} \text{ cm}^3 = 1.664 \times 10^{-23} \text{ cm}^3$.

(D) $1$. In an $fcc$ unit cell, the packing efficiency is $74\%$, which means $74\%$ of the unit cell volume is occupied by atoms.
$2$. The remaining volume is the void volume, which is $100\% - 74\% = 26\%$ of the total unit cell volume $(V_{cell})$.
$3$. Given that the void volume is $1.66 \times 10^{-23} \text{ cm}^3$, we can write: $0.26 \times V_{cell} = 1.66 \times 10^{-23} \text{ cm}^3$.
$4$. Solving for $V_{cell}$: $V_{cell} = \frac{1.66 \times 10^{-23}}{0.26} \text{ cm}^3$.
$5$. $V_{cell} \approx 6.385 \times 10^{-23} \text{ cm}^3$.

(B) In a face-centered cubic $(fcc)$ unit cell, each face particle is located at the center of a face.
Each face of an $fcc$ unit cell is shared by exactly $2$ adjacent unit cells.
Therefore, each face particle is shared by $2$ unit cells.

(B) Step $1$: The mass of one unit cell is given by the product of its volume $(V)$ and its density $(d)$.
Mass of one unit cell = $V \times d = 6.65 \times 10^{-23} \text{ g}$.
Step $2$: The number of unit cells in a given mass of metal is calculated by dividing the total mass of the metal by the mass of one unit cell.
Number of unit cells = $\frac{\text{Total mass}}{\text{Mass of one unit cell}}$.
Step $3$: Substitute the given values:
Number of unit cells = $\frac{0.5 \text{ g}}{6.65 \times 10^{-23} \text{ g}} = 0.07518 \times 10^{23} = 7.518 \times 10^{21}$.
Thus, the correct option is $B$.

(D) Step $1$: In an $hcp$ structure, the number of tetrahedral voids is twice the number of atoms present.
Step $2$: Number of atoms in $0.5 \text{ mole}$ is $0.5 \times N_A = 0.5 \times 6.022 \times 10^{23} = 3.011 \times 10^{23}$.
Step $3$: Number of tetrahedral voids $= 2 \times (\text{Number of atoms}) = 2 \times 3.011 \times 10^{23} = 6.022 \times 10^{23}$.

(C) For an $fcc$ (face-centered cubic) unit cell, the relationship between the edge length $a$ and the atomic radius $r$ is given by: $a = 2\sqrt{2}r$.
Rearranging for $r$: $r = \frac{a}{2\sqrt{2}}$.
Given $a = 408 \text{ pm}$.
Substituting the values: $r = \frac{408}{2 \times 1.414} = \frac{408}{2.828} \approx 144.27 \text{ pm}$.
Rounding to the nearest option, $r = 144.2 \text{ pm}$.

(B) The dissociation reaction for potassium ferrocyanide is:
$K_4[Fe(CN)_6] \rightarrow 4K^+ + [Fe(CN)_6]^{4-}$
Here, the number of ions produced per formula unit is $n = 5$.
The degree of dissociation $(\alpha)$ is $60\% = 0.6$.
The formula for the van't Hoff factor $(i)$ for dissociation is:
$i = 1 + \alpha(n - 1)$
Substituting the values:
$i = 1 + 0.6(5 - 1)$
$i = 1 + 0.6(4)$
$i = 1 + 2.4$
$i = 3.4$