1. Use Raoult's law for vapour pressure lowering

For a non-volatile solute: $P = X_{\text{water}}\, P^0$ where:

• $P^0 = 50\ \text{mm Hg}$ is vapour pressure of pure water at $38^\circ$C,
• $X_{\text{water}}$ is mole fraction of water in the solution.

2. Interpret 1.0 molal solution

A $1.0$ molal solution means:

• $1$ mol of $\mathrm{MgCl_2}$ is dissolved in
• $1000$ g of water.

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

3. Account for dissociation of $\mathrm{MgCl_2}$

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

If fully dissociated, 1 mole gives 3 moles of particles.

Given degree of dissociation: $\alpha = 0.80$

Van't Hoff factor: $i = 1 + (3-1)\alpha = 1 + 2(0.8) = 2.6$

So, effective moles of solute particles: $n_{\text{solute, eff}} = i \times 1 = 2.6$

4. Find mole fraction of water

$X_{\text{water}} = \frac{55.56}{55.56 + 2.6}$

$X_{\text{water}} = \frac{55.56}{58.16} \approx 0.9553$

5. Calculate vapour pressure of solution

$P = X_{\text{water}} P^0 = 0.9553 \times 50$

$P \approx 47.76\ \text{mm Hg}$

Nearest integer: $\boxed{48}$

6. Compare with stored answer

Derived answer is $48$, which matches the stored correct answer.