Question

Dry air was passed through a beaker solution containing $15\text{g}$ of solute and $120\text{g}$ of water. Then, it passed through a beaker containing pure water solvent. Loss in weight of the solution beaker was $2\text{g}$ and loss in weight of solvent beaker was $0.05\text{g}$. What is the molecular weight of the solute ?

• A. $95\text{g/mol}$
• B. $85\text{g/mol}$
• C. $90\text{g/mol}$
• D. $100\text{g/mol}$

Answer 1

The correct option is C $90\text{g/mol}$

Given,
$\text{Loss in mass of the solution beaker}=2\text{g}$
$\text{Loss in mass of the pure solvent}=0.05\text{g}$
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
$or\frac{p^{\circ }-p_s}{p_s}=\frac{0.05}{2}$
For a non-volatile solute,
$p_s=p^{\circ }\times X_{solvent}\therefore \frac{p^{\circ }-p^{\circ }.X_{solvent}}{p^{\circ }.X_{solvent}}=\frac{0.05}{2}\frac{1-X_{solvent}}{X_{solvent}}=\frac{0.05}{2}\frac{X_{solute}}{X_{solvent}}=\frac{0.05}{2}\frac{n_{solute}}{n_{solvent}}=\frac{0.05}{2}$
where,
$n_{solute}$is the number of moles of solute
$n_{solvent}$is the number of moles of solvent
​​$\therefore \frac{n_{solute}}{n_{solvent}}=\frac{\frac{15}{M}}{\frac{120}{18}}=\frac{0.05}{2}M=90\text{g/mol}$