At 20 degree C, the vapor pressures of pure n-propanol and pure isopropanol are

Question

At 20 degree C, the vapor pressures of pure n-propanol and pure isopropanol are 14.6 Torr and 39.0 Torr, respectively. Assuming a solution of the two alcohols is idea, estimate the partial vapor pressure of n-propanol (in torr at 20 degree C) of a solution obtained by combining 200 gm of n-propanol with 100 gm of isopropanol. (The molecular weights are the same for both.) 9.7 14.6 25.9 29.2 What is the total vapor pressure of the solution in the previous problem? 14.6 22.7 30.9 What is the mole fraction of isopropanol in the vapor phase above the solution in problem 3?.333.500.572.667

Explanation / Answer

The number of mols of propanol = 200 g/60. g/mol = 3.33 mol

The number of mols of = 100 g / 60 g/mol = 1.66 mol

Thus - molefraction of propanol = 3.33 mol / (3.33+1.66) = 0.667

          Mole fraction of isopropanol = 1.66 mol / (3.33+1.66) = 0.332

Thus - Partial vapor pressure of n-propanol = (14.6 torr)(0.667) = 9.73 (Answer a)

b) Total Vapor pressure = (14.6 torr)(0.667) + (39 torr)(0.332) = 22.7 torr (answer b)

c) answer is 0.333 (option a)

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