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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Solution:

  1. 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



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