$2.5 \mathrm{~g}$ of a non-volatile, non-electrolyte is dissolved in $100 \mathrm{~g}$ of water at $25^{\circ} \mathrm{C}$. The solution showed a boiling point elevation by $2^{\circ} \mathrm{C}$. Assuming the solute concentration is negligible with respect to the solvent concentration, the vapor pressure of the resulting aqueous solution is _________ $\mathrm{mm}$ of $\mathrm{Hg}$ (nearest integer)

[Given : Molal boiling point elevation constant of water $\left(\mathrm{K}_{\mathrm{b}}\right)=0.52 \mathrm{~K} . \mathrm{kg} \mathrm{mol}^{-1}$, $1 \mathrm{~atm}$ pressure $=760 \mathrm{~mm}$ of $\mathrm{Hg}$, molar mass of water $=18 \mathrm{~g} \mathrm{~mol}^{-1}]$

<p><ul> <li>Molal boiling point elevation constant of water ($K_b$) = 0.52 K.kg.mol^-1</li> <li>1 atm = 760 mm Hg</li> <li>Molar mass of water = 18 g.mol^-1</li> </ul> </p> <p>First, we calculate the molality (m) of the solution using the boiling point elevation formula:</p> <p> <p>$\Delta T_b = K_b \times m$</p> </p> <p>From the problem, we know: <p>$2 = 0.52 \times m$</p> </p> <p>Solving for m: <p>$m = \frac{2}{0.52} \approx 3.846 \text{ mol/kg}$</p> </p> <p>Next, considering the solution is highly diluted, we use the formula for relative lowering of vapor pressure: <p>$\frac{\Delta P}{P^0} = \frac{n_{\text{solute}}}{n_{\text{solvent}}}$</p> </p> <p>Given that molality (m) is defined as the number of moles of solute per kilogram of solvent: <p>$\frac{\Delta P}{P^0} = \frac{m}{1000} \times M_{solvent}$</p> </p> <p>Substitute the known values: <p>$\Delta P = P^0 \times \frac{m}{1000} \times M_{solvent}$</p> <p>$= 760 \times \frac{3.846}{1000} \times 18$</p> <p>$= 52.615 \text{ mm Hg}$</p> </p> <p>Therefore, the vapor pressure of the solution: <p>$P_{solution} = P^0 - \Delta P$</p> <p>$= 760 - 52.615$</p> <p>$\approx 707.385 \text{ mm Hg}$</p> </p> <p>Rounding to the nearest integer, the vapor pressure of the aqueous solution is 707 mm Hg.</p>