Solutions and Colligative Properties - Result Question 37
####8. $18 g$ of 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
(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 $=\frac{\text { Mass }(g)}{\text { Molecular mass }\left(g mol^{-1}\right)}$
$$ =\frac{18 g}{180 gmol^{-1}}=0.1 mol $$
Molar mass of water $=18 g / mol$
Mass of water ( given) $=178.2 g$
Number of moles of water
$$ \begin{aligned} & =\frac{\text { Mass of water }}{\text { Molar mass of water }} \ & =\frac{178.2 g}{18 g / mol}=9.9 mol \end{aligned} $$
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
