Question

The vapour pressure of water at $20^{\circ }C$ is $17.5mmHg$. If $18g$ of glucose $\left(C_6{H}_{12}{O}_6\right)$ is added to $178.2g$ of water at $20^{\circ }C$, the vapour pressure of the resulting solution will be

• A. 17.325 mm Hg
• B. 17.675 mm Hg
• C. 15.750 mm Hg
• D. 16.500 mm Hg

Answer 1

Answer: (A)

In solution containing non-volatile solute, pressure is directly proportional to its mole fraction.

$P_{\text{solution }}=$ vapour pressure of its pure component $\times$
mole fraction in solution

$\therefore P_{\text{sol }}=P^{\circ }{X}_{\text{solvent }}$

Let $A$ be the solute and $B$ the solvent

$\therefore X_B=\frac{n_B}{n_A+n_B}$

$=\frac{\frac{178.2}{18}}{\frac{18}{180}+\frac{178.2}{18}}$

$\Rightarrow X_B=\frac{9.9}{10}=0.99$

Now $P_{\text{solution }}=P^{\circ }{X}_{\text{solvent }}=17.5\times 0.99$

$P_{\text{solution }}=17.32mmHg$