Question

An elevation of $0.70K$ was observed in a solution containing $12.5g$ of non-electrolyte substance in $175g$ of water. Calculate the molar mass of the substance. Molal elevation constant $\left(\right.K_{b}\left.\right)$ for water is $0.52 \, K \, kg \, mol^{- 1}$ .

• A. $53$
• B. $65$
• C. $84$
• D. $79$

Answer 1

Answer: (A)

$\Delta T_{b}=k_{b}\times \frac{w}{M}\times \frac{1000}{W}$
$M=\frac{K_{b} \times w \times 1000}{\Delta T_{b} \times W}$
$M_{B}=\frac{K_{b} \times W_{B} \times 1000}{\Delta T_{b} \times W_{A}}$
According to available data:
Mass of solute $\left(W_{B}\right)=12.5 \, g$
Mass of solute $\left(W_{A}\right)=175 \, g$
Molal elevation constant $\left(K_{b}\right)=0.52 \, K \, kg \, mol^{- 1}$
Elevation in boiling point temperature $\left(\Delta T_{b}\right)=0.70 \, K$
$\therefore M=\frac{\left(0.52 \, K \, k g \, m o l^{- 1}\right) \times \left(12.5 \, g\right) \times 1000}{\left(0.70 \, K\right) \times \left(175 \, g\right)}$
$=53 \, g \, mol^{- 1}$