What mass of urea be dissolved in 171 g of water so as to decrease the vapour pressure of water by 5%?

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25 g

15 g

20 g

30 g

answer is D.

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Detailed Solution

$\frac{P^{°} - P_{s}}{P^{\circ}} = \frac{5}{100} = 0.05 = X_{B} 0.05 = \frac{n_{B}}{n_{B} + n_{A}} = \frac{n_{B}}{n_{B} + \frac{171}{18}} n_{B} = 0.5 weight of urea = 0.5 mol \times 60 g / mol = 30 g$

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