Column $I$	Column $II$
$(A)$. Kohlrausch law can calculate	$(P)$. $\frac{\Lambda_m^c}{\Lambda_m^o}$
$(B)$. Molar conductance $\Lambda_m$	$(Q)$. $\frac{1}{R} \times \frac{l}{A}$
$(C)$. Specific conductance $\kappa$	$(R)$. $\Lambda_m^o$ of $Ca_3(PO_4)_2$
$(D)$. Degree of ionization of weak electrolyte	$(S)$. $\frac{\kappa \times 1000}{M}$

Which of the following options shows the correct matches?

Consider the cell reaction at $300 \ K$: $A_{(s)} + B^{2+}_{(aq)} \rightleftharpoons A^{2+}_{(aq)} + B_{(s)}$. Its $E^{\circ}$ is $1.0 \ V$. The $\Delta_{r}H^{\circ}$ of the reaction is $-163 \ kJ \ mol^{-1}$. What is $\Delta_{r}S^{\circ}$ (in $J \ K^{-1} \ mol^{-1}$) of the reaction? $(F = 96500 \ C \ mol^{-1})$

Consider the following electrochemical cell at standard condition: $Au_{(s)} | QH_2, Q | NH_4X(0.01 \ M) || Ag^{+}(1 \ M) | Ag_{(s)}$. Given $E_{\text{cell}} = +0.4 \ V$. The couple $QH_2 / Q$ represents the quinhydrone electrode,and the half-cell reaction is given as: $Q + 2e^- + 2H^+ \rightarrow QH_2$ with $E^o_{Q/QH_2} = +0.7 \ V$. Given: $E^o_{Ag^+/Ag} = +0.8 \ V$ and $\frac{2.303 \ RT}{F} = 0.06 \ V$. The $pK_b$ value of the ammonium halide salt $(NH_4X)$ used here is $.........$ (nearest integer).

The equation that is incorrect is

The electrochemical cell shown below is a concentration cell.
$M \mid M^{2+} (\text{saturated solution of a sparingly soluble salt, } MX_2) \mid M^{2+} (0.001 \ mol \ dm^{-3}) \mid M$
The emf of the cell depends on the difference in concentration of $M^{2+}$ ions at the two electrodes. The emf of the cell at $298 \ K$ is $0.059 \ V$.
$1.$ The solubility product $(K_{sp}; \ mol^3 \ dm^{-9})$ of $MX_2$ at $298 \ K$ based on the information available for the given concentration cell is (take $2.303 \times R \times 298 / F = 0.059 \ V$):
$(A) \ 1 \times 10^{-15} \quad (B) \ 4 \times 10^{-15}$
$(C) \ 1 \times 10^{-12} \quad (D) \ 4 \times 10^{-12}$
$2.$ The value of $\Delta G \ (kJ \ mol^{-1})$ for the given cell is (take $1 \ F = 96500 \ C \ mol^{-1}$):
$(A) \ -5.7 \quad (B) \ 5.7 \quad (C) \ 11.4 \quad (D) \ -11.4$
Give the answer for question $1$ and $2$.

The molar conductivity of acetic acid solution at infinite dilution is $390 \ S \ cm^2 \ mol^{-1}$. What is the molar conductivity of $0.01 \ M$ acetic acid solution (in $S \ cm^2 \ mol^{-1}$)? (Given: $K_{a}(CH_3COOH) = 1.8 \times 10^{-5}$,assume $1-\alpha \approx 1$)