Question

Certain mass of a non-volatile solute (urea) when dissolved in $100\text{g}$ of water decreased the vapour pressure of water by $30\%$. What is the molality of the solute in the solution?

• A. 30.52 m
• B. 28.20 m
• C. 23.81 m
• D. 35.62 m

Answer 1

The correct option is C 23.81 m

Using Raoult's law,
$\frac{P_o-P_s}{P_o}=x_B=\frac{n_B}{n_A+n_B}$
Here, $x_B$ is mole fraction of solute
$n_A$ and $n_B$ are number of moles of solvent and solute and M and m are masses of solvent and solute.

$\frac{3}{10}=\frac{m/60}{100/18+m/60}$
$m=\frac{1000}{7}$
$Molality=\frac{numberofmoles}{weightofsolute}\times 1000=\frac{\frac{1000/7}{60}}{100}\times 1000$
Molality = 23.81 m