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Relative lowering in vapour pressure of a solution containing 1 mole urea in 54 g $H_{2} O$ is :

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$\frac{3}{4}$

$\frac{1}{55}$

$\frac{3}{55}$

$\frac{1}{4}$

answer is D.

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Detailed Solution

$\frac{P º - P_{s}}{P º} = X_{s o l u t e} = \frac{1}{1 + \frac{54}{18}} = \frac{1}{4}$

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