Question

Boiling point of water at 750 mm Hg is $96.63^{\circ}C$. How much sucrose is to be added to 500 g of water such that it boils at $100^{\circ}C$? Molal elevation constant for water is $0.52 \,K\, kg\, mol^{-1}$.

• A. 11.08 kg
• B. 1108.2 g
• C. 1108.2 kg
• D. 11.08 g

Answer 1

Answer: (B)

Elevation in boiling point $(ΔT_b)$
$= T_{b} - T_{b}^{\circ} = 100 - 96.63 = 3.37^{\circ}C$
Mass of solvent (water), $W_{1} = 500 \,g$
Molar mass of solvent, $M_{1} = 18\, g \,mol^{-1}$
Molar mass of solute, $\left(C_{12}H_{22}O_{11}\right), M_{2} = 342\, g\, mol^{-1}$
Mass of sucrose to be added $\left(W_{2}\right) = ?$
$M_{2} = \frac{1000 \,K_{b} \,W_{2}}{W_{1} \,\Delta T_{b}}$
From formula,
$W_{2} = \frac{M_{2}\times W_{1}\times \Delta T_{b}}{1000\times K_{b} } = \frac{342\times 500\times 3.37}{1000\times 0.52} = 1108.2 \,g$