Question

$18\text{g}$ of glucose $\left(C_6{H}_{12}{O}_6\right)$ is added to $178.2\text{g}$ water at $760$ torr. The vapour pressure of water (in torr) for this aqueous solution is:

• A. $76.0$
• B. $752.4$
• C. $759.0$
• D. $7.6$

Answer 1

The correct option is B $752.4$

Vapour pressure of water $\left(p°\right)=760\text{torr}$
$\text{Number of moles of glucose}=\frac{Mass\left(\text{g}\right)}{Molarmass\left({\text{g mol}}^{-1}\right)}=\frac{18}{180}=0.1\text{mol}$
Mass of water given = $178.2\text{g}$
Number of moles of water = $\frac{\text{Mass of water}}{\text{Molar mass of water}}$
$=\frac{178.2}{18}=9.9\text{mol}$
Total number of moles = $0.1+9.9\text{mol}=10\text{mol}$
Now, mole fraction of glucose in solution = Change in pressure with respect to initial pressure
$i.e.,\frac{\Delta p}{p^o}=\frac{0.1}{10}$
$\Rightarrow \Delta p=0.01p°=0.01\times 760=7.6\text{torr}\therefore$ Vapour pressure of solution $=(760-7.6)\text{torr}=752.4\text{torr}$