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30 gm of urea (M=60 gm /mol) is dissolved in 846 gm of water. Calculate the vapour pressure of water for this solution if the vapour pressure of pure water at 298 K 23.8 mm Hg |
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Answer» Given, weight of urea (W2) = 30g Weight of water (W1) = 846g Vapour pressure of water P1° = 23.8 mm Hg nB = \(\frac{30}{60}\) = 0.5, nA = \(\frac{846}{18}\) = 47 Mole fraction of water (xA) = \(\frac{n_A}{n_A\ +\ n_B}\) =\(\frac{47}{47\ +\ 0.5}\) = \(\frac{47}{47.5}\) = 0.99 PA = PA ° × xA = 23.8 × 0.99 = 23.5 mm Hg. |
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