Question

Vapour pressure in mm $ Hg $ of $ 0.1 $ mole of urea in $ 180\, g $ of water at $ 25^{\circ} C $ is (The vapour pressure of water at $ 25^{\circ} C $ is $ 24\, mm\, Hg) $

• A. 2.376
• B. 20.76
• C. 23.76
• D. 24.76

Answer 1

Answer: (C)

From Raoult's law for very dilute solution.

$\frac{p^{\circ}-p_{s}}{p^{\circ}}=\frac{w}{m} \times \frac{M}{W} $

$\frac{24-p_{s}}{24}=0.1 \times \frac{18}{180} $

$24-p_{s}=0.24 $

$\therefore p_{s}=24-0.24=23.76\, mm\, Hg$