If molar conductance at infinite dilution for $CH_3COO^-$ and $H^+$ ions are $50 \text{ S cm}^2 \text{mol}^{-1}$ and $350 \text{ S cm}^2 \text{mol}^{-1}$ respectively, and the molar conductivity of $5 \times 10^{-2} \text{ M}$ $CH_3COOH$ is $20 \text{ S cm}^2 \text{mol}^{-1}$, what is the hydrogen ion concentration in $\text{mol/dm}^3$ of $CH_3COOH$?

Consider the following statements pertaining to fuel cells :-
$(a)$ Hydrogen-oxygen fuel cells make use of concentrated $KOH$ solution as an electrolyte and porous graphite impregnated with platinum as electrodes.
$(b)$ The efficiency of a fuel cell is less than unity due to polarization at electrodes and the resistance offered by the electrode and the electrolyte.
$(c)$ The electrical work,assuming the cell to be working reversibly,may be represented as $-\Delta G = W_{\text{electrical}} = -\Delta H + T\Delta S$.
Which of the above statements are correct?

Consider the following data.

Electrolyte	$\Lambda_m^\circ$ $(S\text{ cm}^2\text{ mol}^{-1})$
$BaCl_2$	$x_1$
$H_2SO_4$	$x_2$
$HCl$	$x_3$

$BaSO_4$ is sparingly soluble in water. If the conductivity of the saturated $BaSO_4$ solution is $x\text{ S cm}^{-1}$, then the solubility product of $BaSO_4$ can be given as (Here $\Lambda_m = \Lambda_m^\circ$)

Consider an electrochemical cell: $A_{(s)} | A^{n+}(aq, 2 \ M) || B^{2n+}(aq, 1 \ M) | B_{(s)}$. The value of $\Delta H^{\ominus}$ for the cell reaction is twice that of $\Delta G^{\ominus}$ at $300 \ K$. If the emf of the cell is zero,the $\Delta S^{\ominus}$ (in $J \ K^{-1} \ mol^{-1}$) of the cell reaction per mole of $B$ formed at $300 \ K$ is. . . . . . . (Given: $\ln(2) = 0.7, R = 8.3 \ J \ K^{-1} \ mol^{-1}$.)

The conductivity of $0.001028 \, mol \, L^{-1}$ acetic acid is $4.95 \times 10^{-5} \, S \, cm^{-1}$. Calculate its dissociation constant if $\Lambda_m^\circ$ for acetic acid is $390.5 \, S \, cm^2 \, mol^{-1}$.

$A$ battery is constructed of $Cr$ and $Na_2Cr_2O_7$. The unbalanced chemical equation when such a battery discharges is: $Na_2Cr_2O_7 + Cr + H^{+} \to Cr^{3+} + H_2O + Na^{+}$. If one Faraday of electricity is passed through the battery during charging,the number of moles of $Cr^{3+}$ removed from the solution is: