Question

18 g glucose $\left(C_6{H}_{12}{O}_6\right)$ is added to 178.2 g water. The vapor 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

The molar masses of glucose and water are 180 g/mol and 18 g/mol respectively.

Number of moles of glucose $=\frac{18}{180}=0.1$

Number of moles of water $=\frac{178.2}{18}=9.9$

Mole fraction of glucose $=\frac{0.1}{0.1+9.9}=0.01$

Relative lowering in vapour pressure is equal to the mole fraction of glucose.

$\frac{760-P}{760}=0.01$

$760-P=7.6$

$P=760-7.6=752.4$ torr.

Hence, the correct option is $B$