Question

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

Answer 1

`Mw_(2)(C_(2)H_(6)O_(2)) =12 xx 2 + 1 xx 6 + 16 xx 2 =62 g mol^ (-1)`
`W_2(C_(2)H_(6)O_(2))=20 g, W_1 (H_(2) O)=100-20 = 80g `
`n_2(C_(2)H_(6)O_(2))=(20g)/(62g mol^(-1)) = 0.322 mol`
`n_(1)(H_2 O)= (80g) /( 18 g mol ^(1)) = 4.444 mol`
`chi_(glycol) =(n_2)/(n_1+n_2) =( 0.322 mol)/(4.444 + 0.322mol) = 0.068 `
`chi_(H_(2)O)=1-0.068 =0.932`