Question

Vapour pressure of solution containing $6\text{g}$ of a non-volatile solute in $180\text{g}$ of water is $20\text{torr}$. If $1$ mole water is further added, then vapour pressure increases by $0.02\text{torr}$. Which of the following is not correct?

• A. The molecular weight of solute is $55{\text{g mol}}^{-1}$
• B. The vapour pressure of pure water is $20.22\text{torr}$
• C. Addition of more water in the solution will further raise the vapour pressure of the solution
• D. The vapour pressure of pure water is $22.22\text{torr}$

Answer 1

The correct option is D The vapour pressure of pure water is $22.22\text{torr}$

$p_1=$ vapour pressure of solution
$p_1^{\circ }=$vapour pressure of pure solvent
According to Raoult's law,
$\frac{p_1^{\circ }-p_1}{p_1^{\circ }}=\frac{w_2{M}_1}{w_1{M}_2}\Rightarrow \frac{p_1^{\circ }-20}{p_1^{\circ }}=\frac{6\times 18}{M\times 180}=\frac{6\times 1}{M\times 10}.......\left(1\right)$
Again by adding one mole of water, we have
$\frac{p_1^{\circ }-20.02}{p_1^{\circ }}=\frac{6\times 18}{M\times 198}=\frac{6\times 1}{M\times 11}.......\left(2\right)$

Dividing equation $\left(1\right)$ by $\left(2\right)$, we get
$\frac{p_1^{\circ }-20}{p_1^{\circ }}\times \frac{p_1^{\circ }}{p_1^{\circ }-20.02}=\frac{6\times 1}{M\times 10}\times \frac{M\times 11}{6\times 1}\Rightarrow \frac{p_1^{\circ }-20}{p_1^{\circ }-20.02}=\frac{11}{10}\Rightarrow p_1^{\circ }=20.22\text{torr}$

Putting the value of $p_1^{\circ }$ in the equation $\left(2\right)$, we get $M=55.14\approx 55$

On dilution, $\Delta P$ decreases. So, vapour pressure increases.