The vapour pressure of 30% (w/v) aqueous solution of glucose is ____mm Hg at 23^oC
[Given :The density of 30%(w /v) aqueous solution of glucose is 1.2g cm^-3and vapour pressure of pure water is 24mm Hg. [molar mass of glucose is 180gmol^-1] [Round off the answer to nearest integer value]

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answer is 23.

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Detailed Solution

Given, vapour pressure of pure water(P^o) = 24mm Hg

mass of glucose(solute)=30g

volume of sol. = 100ml

mass of sol.= 100 $\times$ 1.2=120g

mass of solvent=120-30=90g

$n_{g l u \cos e} = \frac{30}{180} = 0.17; n_{H_{2} O} = \frac{90}{18} = 5 According to raoult' s law, \frac{p^{o} - p}{p^{o}} = \frac{n_{s}}{n_{o}} \frac{24 - p}{24} = \frac{0.17}{5} p = 23.2 \approx 23 mm of Hg$

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