80 mole percent of MgCl$$_2$$ is dissociated in aqueous solution. The vapour pressure of $$1.0$$ molal aqueous solution of MgCl$$_2$$ at $$38°$$C is _____ mm Hg. (Nearest integer)
Given: Vapour pressure of water at $$38°$$C is $$50$$ mm Hg

For $$MgCl_2$$,

$$MgCl_2 \rightarrow Mg^{2+}+2Cl^-$$

The total number of ions produced is $3$. Therefore, the van't Hoff factor is

$$i=1+\alpha(n-1)$$

$$i=1+0.8(3-1)=2.6$$

A $$1.0$$ molal solution contains $$1$$ mol of $$MgCl_2$$ dissolved in $$1000$$ g of water.

The moles of water are

$$n_{\text{water}}=\frac{1000}{18}=55.56\ \text{mol}$$

Using Raoult's law,

$$\frac{P^\circ-P}{P^\circ}=\frac{i,n_2}{n_1+i,n_2}$$

Substituting the given values,

$$\frac{50-P}{50}=\frac{2.6\times1}{55.56+2.6}=\frac{2.6}{58.16}\approx0.0447$$

Hence,

$$50-P=50\times0.0447\approx2.24\ \text{mm Hg}$$

$$P=50-2.24\approx47.76\ \text{mm Hg}$$

Therefore,

$$P\approx48\ \text{mm Hg}$$

The calculated value of approximately $$2\ \text{mm Hg}$$ represents the lowering of vapor pressure $$(\Delta P)$$, whereas the vapor pressure of the solution is approximately $$48\ \text{mm Hg}$$.