Question

$18\text{g}$ of glucose $(C_6{H}_{12}{O}_6$ is added to $178.2\text{g}$ water. The vapour pressure of water (in torr) for this aqueous solution is: ( Give your answer upto one decimal only)

Answer 1

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