$\text{P}^{\circ}_{\text{A}}=55.3\text{ mm of Hg, P}_{\text{A}}=?$
Mass of Solvent = (100 - 10)g = 90g,
No. of Moles of solvent =$\frac{\text{90g}}{\text{18g mol}^{-1}}=\text{5 mol} $
No. of Moles of solute$\frac{\text{10}}{\text{60}}=\frac{\text{1}}{\text{6}}\text{ mol}$
$\frac{\text{55.3 - P}_{A}}{\text{55.3}}=\frac{\frac{1}{6}}{\frac{1}{6}+5}=\frac{\text{1}}{\text{31}}$
$\text{P}_{A}\text{=53.52 mm Hg}.$
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$\text{Cd}^{2+}(\text{aq})+\text{Zn(s)}\xrightarrow{ \ \ \ \ \ \ \ }\text{Zn}^{2+}(\text{aq) + Cd(s)}$
If
$\text{E}^{\circ}_{\text{Cd}^{2+}/\text{Cd}}=-0.403\text{V;}\text{ E}^{\circ}_{\text{Zn}^{2+}/\text{Zn}}=-0.763\text{V}$