Calculate the vapor pressure lowering when 10.0ml glycerol (C 3 H 8 O 3 ) is add

Question

Calculate the vapor pressure lowering when 10.0ml glycerol (C3H8O3) is added to 500 ml waterat 50'C. At this temperature the vapor pressure of pure water is92.5 torr and its density is 0.988 g/ml. The density of glycerol is12.6 g/ml.

Explanation / Answer

C3H8O3 Molecular weight = 92 10 ml of glycerol , density =12.6g/ml => weight of glycerol =126 g so Moles of glycerol = 126/92 = 1.369 500ml of water of density 0.988g/ml => weight of water = 494 g=0.494 Kg => Using raoult's law: Relative lowering of vapour pressure is dp/p =xs, where xs is the molfraction of solute now xs =1.369/(1.369+(494/18)) = 0.0475 => dp/p = 0.0475 given p = 92.5 torr => dp = 0.0475*92.5 = 4.394torr so Vapour Pressure Lowering = 4. 394tor

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