At $300 \ K$,the $E_{cell}^{\circ}$ of $A_{(s)} + B^{2+}_{(aq)} \rightleftharpoons A^{2+}_{(aq)} + B_{(s)}$ is $1.0 \ V$. If $\Delta_r S^{\circ}$ of this reaction is $100 \ J \ K^{-1} \ mol^{-1}$,what is $\Delta_r H^{\circ}$ (in $kJ \ mol^{-1}$) of this reaction? $(F = 96500 \ C \ mol^{-1})$

If the molar conductivity $(\Lambda_{m})$ of a $0.050 \ mol \ L^{-1}$ solution of a monobasic weak acid is $90 \ S \ cm^{2} \ mol^{-1}$,its extent (degree) of dissociation will be. [Assume $\Lambda_{+}^{\circ} = 349.6 \ S \ cm^{2} \ mol^{-1}$ and $\Lambda_{-}^{\circ} = 50.4 \ S \ cm^{2} \ mol^{-1}$.]

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.

During the electrolysis of carnallite,$MgCl_2$ is decomposed and not $KCl$. This is because of

Which of the following statements is not correct?

For a saturated solution of $Ag_{2}CrO_{4}$ at infinite dilution,$\lambda_{m}^{\infty}(Ag^{+}) = 127 \ \Omega^{-1} \ cm^{2} \ mol^{-1}$ and $\lambda_{m}^{\infty}(CrO_{4}^{2-}) = 246 \ \Omega^{-1} \ cm^{2} \ mol^{-1}$. If the specific conductance of the solution is $2 \times 10^{-2} \ \Omega^{-1} \ cm^{-1}$,calculate the solubility product $(K_{sp})$ of $Ag_{2}CrO_{4}$.