Answer:- \begin{align*} \text{Given:} \\ & \text{Vapour pressure of pure water} = P_1^0 = 17 \text{ mm Hg} \\ & \text{Mass of urea} (W_2) = 2.8 \text{ g} \\ & \text{Mass of water} (W_1) = 50 \text{ g} \\ \text{To find:} \\ & \text{Vapour pressure of the solution} (P_1) \\ \text{Formula:} \\ & \frac{{P_1^0 – P_1}}{{P_1^0}} = \frac{{W_2M_1}}{{M_2W_1}} \\ \text{Calculation:} \\ & \text{Molar mass of urea (NH}_2\text{CONH}_2\text{)} = 14 + 2 + 12 + 16 + 14 + 2 = 60 \text{ g/mol} \\ & \text{Molar mass of water} = 18 \text{ g/mol} \\ & \text{Now, using the formula,} \\ & \frac{{P_1^0 – P_1}}{{P_1^0}} = \frac{{2.8 \text{ g} \cdot 18 \text{ g/mol}}}{{50 \text{ g} \cdot 60 \text{ g/mol}}} \\ & \therefore \frac{{17 \text{ mm Hg} – P_1}}{{17 \text{ mm Hg}}} = 0.0168 \\ & \therefore 17 \text{ mm Hg} – P_1 = 0.0168 \cdot 17 \text{ mm Hg} \\ & \therefore 17 \text{ mm Hg} – P_1 = 0.2856 \text{ mm Hg} \\ & \therefore P_1 = 17 \text{ mm Hg} – 0.2856 \text{ mm Hg} = 16.71 \text{ mm Hg} \\ & \text{The vapour pressure of the given solution is 16.71 mm Hg.} \end{align*}