Question

18 g glucose $\left(C_6{H}_{12}{O}_6\right)$ is added to 178.2 g water. The vapour pressure of water (in torr) for this aqueous
solution is:

• A. 76
• B. 752.4
• C. 759
• D. 7.6

Answer 1

The correct option is B 752.4

Moles of glucose ($C_6{H}_{12}{O}_6)=0.1$

Moles of water $H_{2}O=\frac{178.2}{18}=9.9$
From Raoult's law

$\frac{p^0-p}{p^0}=\frac{0.1}{10}$

$\Rightarrow p^0-p=\frac{0.1}{10}\times 760=7.6$

$\Rightarrow p=p^0-7.6$

$\Rightarrow p=760-7.6$

$=752.4\text{torr}$