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The vapour pressure of water is 12.3 kPa at 300 K. Calculate vapour pressure of 1 molal solution of a non-volatile solute in it. |
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Answer» P° H2O = 12.3 kPa 1000 In 1 molal solution, nsolute = 1; nH2O = \(\frac{1000}{8}\) = 55.5 ∴ XH2O = \(\frac{55.5}{55.5\,+\,1}\) Vapour pressure of the solution, Ps = P°H2O × XH2O = 0.982 × 12.3 = 12.08 kPa |
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