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A glucose solution which boils at 101.04°C at 1 atm. What will be relative lowering of vapour pressure of an aqueous solution of urea which is equimolal to given glucose solution? (Given: Kb for water is 0.52 K kg mol-1 ) |
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Answer» ΔTb = Kb m ΔTb = 101.04 - 100 = 1.04 oC or m= \(\frac{1.04}{0.52}\) = 2 m 2 m solution means 2 moles of solute in 1 kg of solvent. 2 m aq solution of urea means 2 moles of urea in 1kg of water. No. of moles of water = \(\frac{1000}{18}\) = 55.5 Relative lowering of VP = x2 (where x2 is mole fraction of solute) Relative lowering of VP = \(\frac{n_2}{n_1}+n_2\) (n2 is no. of moles of solute , n1 is no. of moles of solvent) = \(\frac{2}{2}\)+55.5 = \(\frac{2}{57.7}\) = 0.034 |
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