Question

A solution containing $30\text{g}$ of a non-volatile solute in exactly $90\text{g}$ of water has a vapour pressure of $21.85\text{mm of Hg}$ at ${25}^{o}C.$ Further, $18\text{g}$ of water is then added to the solution, the new vapour pressure becomes $22.15\text{mm of Hg}$ at ${25}^{o}C.$ Calculate:
(i) Molar mass of the solute (in g/mol) and
(ii) Vapour pressure of water at ${25}^{o}C.$

• A. $\left(i\right)=23.87,\left(ii\right)=97\text{mm Hg}$
• B. $\left(i\right)=87,\left(ii\right)=67.9\text{mm Hg}$
• C. $\left(i\right)=67.9,\left(ii\right)=23.87\text{mm Hg}$
• D. $\left(i\right)=23.87,\left(ii\right)=67.9\text{mm Hg}$

Answer 1

The correct option is C $\left(i\right)=67.9,\left(ii\right)=23.87\text{mm Hg}$

Let the vapour pressure of water at ${25}^{o}C$ be $p_0$ and molecular mass of the solute be $m$.
Using Raoult's law in the following form,
$\frac{p_0-p_s}{p_s}=\frac{wM}{Wm}$
Where,
$p_s=\text{Vapour pressure of solution}$
$w=\text{mass of solute}W=\text{mass of solvent}M=\text{Molar mass of solvent}$
For solution $\left(I\right)$,
Substituting the values, we get,
$\frac{(p_0-21.85)}{21.85}=\frac{30\times 18}{90\times m}...\left(i\right)$
For solution $\left(II\right)$,
$\frac{(p_0-22.15)}{22.15}=\frac{30\times 18}{108\times m}...\left(ii\right)$

Dividing eq. (i) by eq. (ii),
$\frac{(p_0-21.85)}{21.85}\times \frac{22.15}{(p_0-22.15)}=\frac{108}{90}=\frac{6}{5}$

$p_0=23.87\text{mm of Hg}$
Substituting the value of $p_0$ in equation $\left(i\right)$
$m=67.9g/mol$