Question

The vapour pressure of water at ${20}^{o}C$ 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 ${20}^{o}C$, the vapour pressure of the resulting solution will be :

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

Answer 1

The correct option is B 17.325 mm Hg

Relative lowering of vapour pressure:

The vapour pressure of a liquid is the pressure of the vapour which is in equilibrium with that liquid. The vapour pressure of a solvent is lowered when a non-volatile solute is dissolved in it to form a solution. The reduction of the vapor pressure of the solvent is given as

$p_1=x.{p_1}^0$

$p_1$ is the vapor pressure of the solution.

${p_1}^0$ is the vapor pressure of pure solvent

$x$ is the mole fraction of the solvent.

Vapour pressure of pure water (solvent) $=17.5$ mm of Hg

Mass of glucose (solute) $=18$ g

No. of moles of solute $=\frac{18}{180}=0.1$ mol

Mass of water $=178.2$ g

No. of moles of water $=\frac{178.2}{18}=9.9$ mol

Mole fraction of solvent will be $\frac{9.9}{(9.9+0.1)}=0.99$

Vapour pressure of the solution is given by

$p=x.p^0$

$p=0.99\times 17.5$

$p=17.325$ mm of Hg vapor pressure of the solution.

Hence, the correct option is $\text{B}$