Question

An aqueous solution of $2$% non-volatile solute exerts a pressure of $1.004$ bar at the normal boiling point of the solvent. What is the molecular mass of the solute?

• A. $23.4$ $g{mol}^{-1}$
• B. $41.35$ $g{mol}^{-1}$
• C. $10$ $g{mol}^{-1}$
• D. $20.8$ $g{mol}^{-1}$

Answer 1

Answer: (B)

$41.35$ $g{mol}^{-1}$

At boiling point, the vapour pressure of pure water is $1$ atm or $1.013$ bar.

Vapour pressure of the solution is $1.004$ bar.

Mass of solute is $2$ g.

Mass of solution is $100$ g.

Mass of solvent is $100-2=98$ g

Applying Raoult's law, we get

$\frac{P^o-P}{P}=x_2=\frac{W_2{M}_1}{M_2{W}_1}$

$\frac{1.013-1.004}{1.013}=\frac{2\times 18}{M_2\times 98}$ $⟹M_2=41.35$ g/mol

Hence, the correct option is $B$