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Find the vapour pressure of solution containing 5g of sucrose in 1kg of water. If the vapour pressure of water is 4.7 mm of Hg. |
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Answer» As we know, relative lowering of vapour pressure \(\frac{P^0- P_s}{P^0} = x_2\) .....(1) where, x2 = mole fraction of solute P0 = pure vapour pressure of water Ps = vapour pressure of solution \(\therefore\) Number of moles of sucrose = \(\frac 5{342} = 0.1444\) Number of moles of water = \(\frac{1000}{18}= 55.56\) \(\therefore\) Mole fraction of sucrose = \(\frac{0.0144}{0.0144 + 55.56} = 0.0003\) \(\therefore\) From equation (1) \(\frac {4.7 - P_s}{4.7} = 0.0003\) \(4.7 - P_s = 0.00141\) \(P_s = 4.6986 \, mm\,Hg\) \(\therefore\) Vapour pressure of solution will be 4.6986 mm Hg. |
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