- A$S$ : ground state, covalency $= 6$
- B$S$ : excited state, covalency $= 4$
- ✓$S$ : excited state, oxidation No $= + 6$
- D$S$ : ground state, oxidation No. $= + 6$

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(Assume : Densities of the solutions $A$ and $B$ are the same as that of water and the soluble salts dissociate completely.)
Use: Molal elevation constant (Ebullioscopic Constant), $K _{ b }=0.5 K kg mol ^{-1}$; Boiling point of pure water as $100^{\circ} C$.)
($1$) The value of $x$ is
($2$) The value of $|y|$ is

$\mathrm{M}(\mathrm{s}) \mid \mathrm{M}^{+}$(aq; $0.05$ molar) || $\mathrm{M}^{+}(\mathrm{aq}), 1$ molar) $\mid \mathrm{M}(\mathrm{s})$
For the above electrolytic cell the magnitude of the cell potential $\left|E_{\text {cell }}\right|=70 \mathrm{mV}$.
$1.$ For the above cell
$(A)$ $\mathrm{E}_{\text {cell }}<0 ; \Delta \mathrm{G}>0$ $(B)$ $\mathrm{E}_{\text {cell }}>0 ; \Delta \mathrm{G}<0$
$(C)$ $\mathrm{E}_{\text {cell }}<0 ; \Delta \mathrm{G}^{\circ}>0$ $(D)$ $\mathrm{E}_{\text {cell }}>0 ; \Delta \mathrm{G}^{\circ}>0$
$2.$ If the $0.05$ molar solution of $\mathrm{M}^{+}$is replaced by $0.0025$ molar $\mathrm{M}^{+}$solution, then the magnitude of the cell potential would be
$(A)$ $35 \mathrm{mV}$ $(B)$ $70 \mathrm{mV}$ $(C)$ $140 \mathrm{mV}$ $(D)$ $700 \mathrm{mV}$
Give the answer question $1,2.$