Question

Dry air was passed through a solution containing $20g$ of a substance in $100.0g$ of water and then through pure water. The loss in mass of the solution was $3g$ and that of pure water was $0.06g$. Calculate the molar mass of the substance.

• A. $198.2g/mol$
• B. $180g/mol$
• C. $108.2g/mol$
• D. $208.6g/mol$

Answer 1

The correct option is B $180g/mol$

Given,
$\text{Loss in mass of the solution}=3g$
$\text{Loss in mass of the pure solvent}=0.06g$
By Ostwald-Walker method,
We have,
$\frac{\text{loss in mass of pure water}}{\text{loss in mass of solution}}=\frac{p^{\circ }-p_s}{p_s}$
where,
$p^{\circ }$ is pure vapour pressure of solvent
$p_s$ is vapour pressure of the solution
$or\frac{p^{\circ }-p_s}{p_s}=\frac{0.06}{3}=0.02$
By relative lowering of vapour pressure,
$\frac{p^{\circ }-p_s}{p_s}=\frac{n}{N}$
where,
n is number of moles of solute
N is number of moles of solvent
$\therefore \frac{n}{N}=\frac{\frac{20}{M}}{\frac{100}{18}}=0.02$
where,
M is the molar mass of solute
$\therefore$
$M=180g/mol$