Solutions and Colligative Properties 1 Question 7
8. $18 \mathrm{~g}$ of glucose $\left(\mathrm{C}{6} \mathrm{H}{12} \mathrm{O}_{6}\right)$ is added to $178.2 \mathrm{~g}$ water. The vapour pressure of water (in torr) for this aqueous solution is
(2016 Main)
(a) 76.0
(b) 752.4
(c) 759.0
(d) 7.6
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Solution:
- Key Idea Vapour pressure of water $\left(p^{\circ}\right)=760$ torr
Number of moles of glucose $=\frac{\text { Mass }(\mathrm{g})}{\text { Molecular mass }\left(\mathrm{g} \mathrm{mol}^{-1}\right)}$
$$ =\frac{18 \mathrm{~g}}{180 \mathrm{gmol}^{-1}}=0.1 \mathrm{~mol} $$
Molar mass of water $=18 \mathrm{~g} / \mathrm{mol}$
Mass of water ( given) $=178.2 \mathrm{~g}$
Number of moles of water
$$ \begin{aligned} & =\frac{\text { Mass of water }}{\text { Molar mass of water }} \ & =\frac{178.2 \mathrm{~g}}{18 \mathrm{~g} / \mathrm{mol}}=9.9 \mathrm{~mol} \end{aligned} $$
Total number of moles $=(0.1+9.9)$ moles $=10$ moles
Now, mole fraction of glucose in solution $=$ Change in pressure with respect to initial pressure
$$ \begin{aligned} & \text { i.e. } \quad \frac{\Delta p}{p^{\circ}}=\frac{0.1}{10} \ & \text { or } \quad \Delta p=0.01 p^{\circ}=0.01 \times 760=7.6 \text { torr } \end{aligned} $$
$\therefore$ Vapour pressure of solution $=(760-7.6)$ torr $=752.4$ torr
