1. Use boiling point elevation to find molality

For a non-volatile, non-electrolyte:

$\Delta T_b = K_b m$

Given:

$\Delta T_b = 2\,\text{K}, \qquad K_b = 0.52\,\text{K kg mol}^{-1}$

So,

$m = \frac{\Delta T_b}{K_b} = \frac{2}{0.52} = 3.846\,\text{mol kg}^{-1}$

2. Calculate moles of solute in 100 g water

Mass of water:

$100\,\text{g} = 0.100\,\text{kg}$

Hence moles of solute:

$n_2 = m \times 0.100 = 3.846 \times 0.100 = 0.3846\,\text{mol}$

3. Calculate moles of water

$n_1 = \frac{100}{18} = 5.556\,\text{mol}$

4. Find mole fraction of solvent

For a non-volatile solute, by Raoult’s law:

$P = X_1 P^0$

Since the question says solute concentration is negligible with respect to solvent concentration, we use the approximation:

$X_1 \approx \frac{n_1}{n_1+n_2}$

Now,

$X_1 = \frac{5.556}{5.556+0.3846} = \frac{5.556}{5.9406} \approx 0.9353$

5. Vapour pressure of solution

Taking vapour pressure of pure water as $1\,\text{atm} = 760\,\text{mm Hg}$,

$P = 0.9353 \times 760 \approx 710.8\,\text{mm Hg}$

Nearest integer:

$P \approx 711\,\text{mm Hg}$

6. Compare with stored answer

Derived answer is 711 mm Hg, whereas stored correct answer is 707 mm Hg.

The likely reason for the discrepancy is that the exact data used here lead to about $711\,\text{mm Hg}$. The value $707$ does not follow from the standard application of boiling point elevation and Raoult’s law with the given constants.

If one uses the approximation for dilute solution directly:

$\frac{\Delta P}{P^0} = X_2 \approx \frac{n_2}{n_1} = \frac{0.3846}{5.556} \approx 0.0692$

Then,

$P = 760(1-0.0692) \approx 707.4\,\text{mm Hg}$

which gives 707 mm Hg.

Since the problem explicitly says solute concentration is negligible compared to solvent concentration, this dilute-solution approximation is intended. Hence the exam-style expected answer is 707 mm Hg.