80 mole percent of MgCl 2 is dissociated in aqueous solution. The vapour pressure of 1 . 0 molal aqueous…

One molal solution means one of solute present in one kg of solvent.

$$\begin{gathered} {\mathrm{MgCl}}_2\stackrel{}{\rightleftharpoons }{\mathrm{Mg}}^{2+}+2{\mathrm{Cl}}^{-} \\ 1 0 0 \\ 1-0.8 0.8 1.6 \end{gathered}$$

Hence, overall moles after dissociation $=2.6$

The relative lowering of vapour pressure

$\frac{{\mathrm{p}}^{\mathrm{o}}-{\mathrm{p}}_{\mathrm{s}}}{{\mathrm{p}}^{\mathrm{o}}}=\frac{\mathrm{moles} \mathrm{of} \mathrm{solute}}{\mathrm{moles} \mathrm{of} \mathrm{solute}+\mathrm{moles} \mathrm{of} \mathrm{solvent}}$

$\frac{\mathrm{p}°-{\mathrm{p}}_{\mathrm{s}}}{\mathrm{p}°}=\frac{2.6}{{\displaystyle \frac{1000}{18}+2.6}}$

For dilute solution

$\frac{\mathrm{p}°-{\mathrm{p}}_{\mathrm{s}}}{\mathrm{p}°}=\frac{2.6}{{\displaystyle \frac{1000}{18}}}$

$\mathrm{p}=47.66\simeq 48 \mathrm{mm} \mathrm{Hg}$