An aqueous solution containing $6.5 \ g$ of $NaCl$ of $90 \%$ purity was subjected to electrolysis. After the complete electrolysis,the solution was evaporated to get solid $NaOH$. The volume of $1 \ M$ acetic acid required to neutralize $NaOH$ obtained above is (in $cm^{3}$)

Consider the strong electrolytes $Z_{m}X_{n}$,$U_{m}Y_{p}$ and $V_{m}X_{n}$. Limiting molar conductivity $(\Lambda^0)$ of $U_{m}Y_{p}$ and $V_{m}X_{n}$ are $250 \ S \ cm^2 \ mol^{-1}$ and $440 \ S \ cm^2 \ mol^{-1}$,respectively. The value of $(m + n + p)$ is . . . . . Given:

$Ion$	$\lambda^0 \ (S \ cm^2 \ mol^{-1})$
$U^{p+}$	$50.0$
$Y^{m-}$	$50.0$
$V^{n+}$	$60.0$
$X^{m-}$	$50.0$
$Z^{n+}$	$40.0$

$\lambda^0$ is the limiting molar conductivity of ions. The plot of molar conductivity $(\Lambda)$ of $Z_{m}X_{n}$ $vs$ $c^{1/2}$ is given below.

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

Given:
$\lambda _{ClCH_2COONa} = 224 \ \Omega ^{-1} \ cm^2 \ gmeq^{-1}$
$\lambda _{NaCl} = 38.2 \ \Omega ^{-1} \ cm^2 \ gmeq^{-1}$
$\lambda _{HCl} = 203 \ \Omega ^{-1} \ cm^2 \ gmeq^{-1}$
What is the value of $\lambda _{ClCH_2COOH}$ in $\Omega ^{-1} \ cm^2 \ gmeq^{-1}$?

Which one of the following statements is correct?

Given ${E^o}_{Hg^{2+}|Hg_2^{2+}} = 0.9 \ V$ and ${E^o}_{Hg_2^{2+}|Hg} = 0.8 \ V$. Calculate the value of ${\Delta _r}{G^o}$ at $25 \ ^oC$ in $kJ/mol$ for the reaction: $Hg_2^{2+} \to Hg^{2+} + Hg_{(l)}$