Solutions and Colligative Properties - Result Question 7
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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Answer:
Correct Answer: 8. (b)
Solution:
Key Idea: Vapour pressure of water $\left(p^{\circ}\right)=760$ torr
$\text { Number of moles of glucose }=\frac{\text { Mass }(\mathrm{g})}{\text { Molecular mass }\left(\mathrm{g} \mathrm{mol}^{-1}\right)} =\frac{18 g}{180 g mol^{-1}}=0.1 mol$
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
BETA
