Question

18 g of glucose $\left(C_6{H}_{12}{O}_6\right)$ is added to 178.2 g of water. The vapour pressure of water for this aqueous solution at ${100}^{\circ }C$ is

• A. 759.00 torr
• B. 7.60 torr
• C. 76.00 torr
• D. 752.40 torr.

Answer 1

The correct option is D 752.40 torr.

Relative lowering of vapour pressure is equal to mole fraction of glucose
$\frac{P^o-P}{P^o}=\text{mole fraction of glucose}...\left(i\right)\text{moles of glucose}=\frac{18}{180}=0.1\text{moles of water}=\frac{178.2}{18}=9.9\text{mole fraction of glucose}=\frac{0.1}{0.1+9.9}=0.01$
Substituting in equation (i) we get,
$\frac{760-P}{760}=0.01$
On solving, we get,
$P=752.4torr$