The vapour pressure of 30% (w/v) aqueous solution of glucose is ______ mm Hg at 25°C. [Given: The density of 30% (w/v) aqueous solution of glucose is 1.2 g and vapour pressure of pure water is 24 mm Hg. Molar mass of glucose is 180 g .] [20

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Solution

Q.
The vapour pressure of 30% (w/v) aqueous solution of glucose is ______ mm Hg at 25°C.

[Given: The density of 30% (w/v) aqueous solution of glucose is 1.2 g ${cm}^{- 3}$ and vapour pressure of pure water is 24 mm Hg. Molar mass of glucose is 180 g ${mol}^{- 1}$ .] [2023]

Ans.
(23)

$\frac{P^{0} - P_{s}}{P_{s}} = \frac{n}{N}$ ...(i)

$density of solution = \frac{Mass}{volume}$

$Volume of solution = 100 ml$

$Mass = 120 g$

$weight of glucose = 120 \times \frac{30}{100} = 36 g$

${weight of H}_{2} O = 120 - 36 = 84 g$

$mole of glucose = \frac{36}{180} = 0.2 mole$

${mole of H}_{2} O = \frac{84}{18} = 4.6 mole$

So, from eqn (i),

$\frac{24 - P_{s}}{P_{s}} = \frac{0.2}{4.67}$

$24 - P_{s} = 0.0428 P_{s}$

$P_{s} = \frac{24}{1.0428} = 23.015 mm of Hg \approx 23 mm of Hg$

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