The equilibrium constant of a $2$ electron redox reaction at $298 \, K$ is $3.8 \times 10^{-3}$. The cell potential $E^{\circ}$ (in $V$) and the free energy change $\Delta G^{\circ}$ (in $kJ \, mol^{-1}$) for this equilibrium,respectively are

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? $(F = 96500 \ C \ mol^{-1})$

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 $1-\alpha \approx 1$)

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?

The conductivity of $0.01 \ M$ aqueous acetic acid measured with a conductivity cell of cell constant $0.5 \ cm^{-1}$ at $298 \ K$ is $3.12 \times 10^{-4} \ S$. If the limiting molar conductivities of $H^{+}$ and $CH_3COO^{-}$ at the same temperature are $349$ and $41 \ S \ cm^2 \ mol^{-1}$ respectively,the dissociation constant of acetic acid is

All the energy released from the reaction $X \rightarrow Y, \Delta_{r}G^0 = -193 \ kJ \ mol^{-1}$ is used for oxidizing $M^{+}$ as $M^{+} \rightarrow M^{3+} + 2e^-, E^0 = -0.25 \ V$. Under standard conditions,the number of moles of $M^{+}$ oxidized when one mole of $X$ is converted to $Y$ is $\left[F = 96500 \ C \ mol^{-1}\right]$.