80 mole percent of MgCl _ 2 is dissociated in aqueous solution. The...

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

${MgCl}_{2}\overset{}{⇌}{Mg}^{2 +} + 2{Cl}^{-}1001 - 0 . 80 . 81 . 6$

Hence, overall moles after dissociation $= 2 . 6$

The relative lowering of vapour pressure

$\dfrac{{p}^{o} - {p}_{s}}{{p}^{o}} = \dfrac{molesofsolute}{molesofsolute + molesofsolvent}$

$\dfrac{p ° - {p}_{s}}{p °} = \dfrac{2 . 6}{\dfrac{1000}{18} + 2 . 6}$

For dilute solution

$\dfrac{p ° - {p}_{s}}{p °} = \dfrac{2 . 6}{\dfrac{1000}{18}}$

$p = 47 . 66 ≃ 48mmHg$