Boiling point of water at $750 \, mm \, Hg$ is $99.63^{\circ} \, C$. How much sucrose is to be added to $500 \, g$ of water such that it boils at $100^{\circ} \, C$?

(D) The elevation in boiling point is $\Delta T_{b} = 100^{\circ} \, C - 99.63^{\circ} \, C = 0.37 \, K$.
Mass of solvent $(w_{1}) = 500 \, g$.
Molar mass of sucrose $(C_{12}H_{22}O_{11}) (M_{2}) = 342 \, g \, mol^{-1}$.
Molal elevation constant $(K_{b})$ for water is $0.52 \, K \, kg \, mol^{-1}$.
Using the formula $\Delta T_{b} = \frac{K_{b} \times 1000 \times w_{2}}{M_{2} \times w_{1}}$,we get:
$w_{2} = \frac{\Delta T_{b} \times M_{2} \times w_{1}}{K_{b} \times 1000} = \frac{0.37 \times 342 \times 500}{0.52 \times 1000} \approx 121.67 \, g$.
Thus,$121.67 \, g$ of sucrose must be added.