Question

The vapour pressure of water at $20^{\circ} C$ is $17.5\, mm\, Hg$. If $18\, g$ of glucose $\left( C _{6} H _{12} O _{6}\right)$ is added to $178.2 \,g$ 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 $P_{\text {solution }}=P^{\circ} X_{\text {solvent }}=17.5 \times 0.99$

$P_{\text {solution }}=17.32\, mm \,Hg$