The molar conductivity of 0.02 M solution of an electrolyte is $124 \times 10^{-4} \mathrm{~S} \mathrm{~m}^2 \mathrm{~mol}^{-1}$. What is the resistance of same solution (in ohms), kept in a cell of cell constant $129 \mathrm{~m}^{-1}$ ?

If the degree of dissociation of formic acid is $11.0 \%$, the molar conductivity of 0.02 M solution of it is (Given, $\lambda^{\circ}\left(\mathrm{H}^{+}\right)=349.6 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$ $\lambda^{\circ}\left(\mathrm{HCOO}^{-}\right)=54.6 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$ )

Identify the correct statements from the following

(A) At 298 K , the potential of hydrogen electrotle placed in a solution of $\mathrm{pH}=10$, is -0.59 V

(B) The limiting molar conductivity of $\mathrm{Ca}^{2+}$ and $\mathrm{Cl}^{-}$is 119 and $76 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$ respectively. The limiting molar conductivity of $\mathrm{CaCl}_2$ is $195 \mathrm{Scm}^2 \mathrm{~mol}^{-1}$

(C) The correct relationship between $K_C$ and $E_{\text {cell }}^{\ominus}$ is $$ E_{\text {cell }}^\theta=\frac{2303 R T}{n F} \log K_C $$

The reduction potential of a half-cell consisting of a Pt electrode immersed in $2.0 \mathrm{M} \mathrm{Fe}^{2+}$ and $0.02 \mathrm{M} \mathrm{Fe}^{3+}$ solution (in V) is

Given : $\left(\frac{2.303 R T}{F}=0.059, E_{\mathrm{Fe}^{3+} \mid \mathrm{Fe}^{2+}}^{\circ}=0.771 \mathrm{~V}\right)$