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The relative lowering of vapour pressure of a solution containing 6 g of urea dissolved in 90 g of water is

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O.05

1.50

0.01

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detailed solution

Correct option is A

Wt. of urea $= W_{1} = 6 g$

GMW of urea $= M_{1} = 60 g$

Wt of water $= W_{2} = 90 g$

GMW of water $= M_{2} = 18 g$

According to Raoult’s law.

$R . L . V . P = \frac{W_{1}}{M_{1}} \times \frac{M_{2}}{W_{2}}$

$= \frac{6}{60} \times \frac{18}{90}$

$= \frac{1}{50}$

=0.02(Near by)

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