Question

A solution is 25% H2O, 25% ethanol and 50% acetic acid by mass .Calculate the mole fraction of ethanol and acetic acid in the solution

Answer 1

We are given that the solution contains 25% water, 25% ethyl alcohol and 50 % water.

So in 100g of the solution

Mass of water = 25 g

Mass of ethanol = 25 g

Mass of acetic acid = 50 g

Moles of water = n1 = 25/18 =1.39 molesMoles of C2H5OH = n2 = 25/46 = 0.54 molesMoles of CH3COOH= n3 = 50/60 = 0.83 moles

Total moles in solution = n1 + n2 + n3 = 1.39 + 0.54 + 0.83 = 2.76 moles

Mole fraction of water = n1/(n1+n2+n3) = 1.39/2.76 = 0.503

Mole fraction of C2H5OH = n2/(n1+n2+n3) = 0.54/2.76 = 0.196

Mole fraction of CH3COOH = n3/(n1+n2+n3) = 0.83/2.76 = 0.301