Question

Calculate the mole fraction of ethylene glacol `(C_(2)H_(6)O_(2))` in a solution containing 20% of ethylene glycol by mass.

Answer 1

Let us start with 100 g of the solution in which
Mass of water= 80.0 g
Molar mass of ethylene glycol `(C_(2)H_(6)O_(2))=2xx12+6xx1+2xx16=62 g mol^(-1)`
`n_(C_(2)H_(6)O_(2))=((20g))/((62gmol^(-1)))=0.322 mol`
Molar mass of water `(H_(2)O)=2xx1+1xx16=18mol^(-1)`
`n_(H_(2)O)=((80g))/((18gmol^(-1)))=4.444mol`
`x_(C_(2)H_(6)O_(2))=x_(C_(2)H_(6)O_(2))/(x_(C_(2)H_(6)O_(2))+n_(H_(2)O))=(0.322mol)/((0.322mol)+(4.444mol))=((0.322mol))/((4.766mol))=0.0678`