Question

At 35.0degrees, the vapour pressure of pure ethanol (C2H5OH) is 100.0mmHg and the vapour pressure of pure acetone (CH3COCH3) is 360.0 mmHg

a) a solution is formed from 25.0g of ethanol and 15.0g of acetone. Assume ideal behaviour and calculate the vapour pressure of each component above this solution.

b) calculate the mole fraction of acetone in the vapour pressure in equilibrium with the solution in part a).

c) Name a method that can be used to separate these two components in the lab.

Answer 1

Molecular wt of ethanol = 46

So, 25.0 g of ehanol = 25/46= 0.543 mole

Molecular weight of acetone = 58

So, 15 g of acetone = 15/58 = 0.259 mole

total mol of mixture = 0.543+0.259 = 0.802 mole

mole fraction of ethanol, X_ethanol= 0.543/0.802= 0.6771

mole fraction of acetone, X_acetone= 0.259/0.802= 0.3229 [Answer b]

Partial vapour pressure of ethanol, P_ethanol= X_ethanol X P⁰_ethanol = 0.6771x 100 = 67.71 mm of Hg [P⁰_ethanol is vapour pressure of pure ethanol]

Partial vapour pressure of acetone, P_acetone= X_acetone X P⁰_acetone = 0.3229x 360 = 116.244 mm of Hg [P⁰_acetone is vapour pressure of pure acetone] [Answer a]

c) The laboratory method by which the two volatile liqiud can be seperated is fractional distillation, as they have good difference in their boiling point.