Question

A $5$ percent aqueous solution by mass of a nonvolatile solute boils at ${100.13}^{\circ }\text{C}$. Calculate the molar mass of the solute.
(Given : $K_b=0.52{\text{K kg mol}}^{-1}$)

• A. $200.5{\text{g mol}}^{-1}$
• B. $210.5{\text{g mol}}^{-1}$
• C. $100.25{\text{g mol}}^{-1}$
• D. $105.25{\text{g mol}}^{-1}$

Answer 1

The correct option is B $210.5{\text{g mol}}^{-1}$

Elevation in boiling point is
$\Delta T_b=373.13\text{K}-373\text{K}=0.13\text{K}$
Molality of the solution is
$m=\frac{\Delta T_b}{K_b}=\frac{0.13}{0.52}=0.25{\text{mol kg}}^{-1}$
The solution contains $5\%$ by mass of solute, i.e. $1000\text{g}$ of the solution contains $50\text{g}$ of the solute.
Mass of the solvent $=950\text{g}=0.950\text{kg}$
We know that molality,
$m=\frac{w_2}{M_2}\times \frac{1}{w_1}\Rightarrow 0.25=\frac{0.05}{M_2}\times \frac{1}{0.95}\Rightarrow M_2=\frac{0.05}{0.25\times 0.95}=0.2105{\text{kg mol}}^{-1}=210.5{\text{kg mol}}^{-1}$