Question

The boiling point of a solution of $0.11 \,g$ of a substance in $15 \,g$ of ether was found to be $0.1^{\circ} C$ higher than that of pure ether. The molecular weight of the substance will be $\left(K_{b}=2.16^{\circ} C \,kg\, mol ^{-1}\right)$

• A. 148
• B. 158
• C. 168
• D. 178

Answer 1

Answer: (B)

$\Delta T_{b}=\frac{w_{2} \times 1000 \times K_{b}}{M_{2} \times w_{1}}$

$M_{2}=\frac{1000 K_{b} w_{2}}{w_{1} \Delta T_{b}}=\frac{1000 \times 2.16 \times 0.11}{15 \times 0.1}$

$=158.4 \approx 158$