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Solutions Part 1

Section: Chemistry

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Question : 84 of 100

Marks: +1, -0

The vapour pressure of water at

20^{\circ} \mathrm{C}

17.5 \mathrm{mm} \mathrm{Hg}

18 \mathrm{g}

(\mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6)

178.2 \mathrm{g}

20^{\circ} \mathrm{C}

, the vapour pressure of the resulting solution will be

$17.675 \mathrm{mm} \mathrm{Hg}$
$15.750 \mathrm{mm} \mathrm{Hg}$
$16.500 \mathrm{mm} \mathrm{Hg}$
$17.325 \mathrm{mm} \mathrm{Hg}$

Solution:

NOTE : On addition of glucose to water, vapour pressure of water will decrease. The vapour pressure of a solution of glucose in water can be calculated using the relation

\frac{P^{\circ}-P_S}{P_S} = \frac{\text{ Moles of glucose in solution }}{\text{ moles of water in solution }}

or

\frac{17.5-P_S}{P_S} = \frac{\frac{18}{180}}{\frac{178.2}{18}}

[as

P^{\circ}=17.5

]

or

17.5-P_S = \frac{0.1 \times P_S}{9.9}

or

P_S=17.325 \mathrm{mm} \mathrm{Hg}

.

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