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 $(K_b)$ for water is $0.52Kkgmo{l}^{-1}$ .

• 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 $\left(W_B\right)=12.5g$
Mass of solute $\left(W_A\right)=175g$
Molal elevation constant $\left(K_b\right)=0.52Kkgmo{l}^{-1}$
Elevation in boiling point temperature $\left(\Delta T_b\right)=0.70K$
$\therefore M=\frac{\left(0.52Kkgmo{l}^{-1}\right)\times \left(12.5g\right)\times 1000}{\left(0.70K\right)\times \left(175g\right)}$
$=53gmo{l}^{-1}$