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.

Options

A. `0.92`
B. `0.76`
C. `0.82`
D. `0.36`

Answer

A

Explanation

`"Molecular mass of "C_(2)H_(2)O_(2)`
`=12xx2+1xx6+16xx2=62g" mol"^(-1)`
`w_(2)(C_(2)H_(6)O_(2))=20g,w_(1)(H_(2)O)=100-20=80g`
`n_(2)(C_(2)H_(6)O_2)=(20g)/(60g" mol"^(-1))=0.322mol`
`n_(1)(H_(2)O)(80g)/(18g" mol"^(-1))=4.444mol`
`x_("glycol")=(n_2)/(n_(1)+n_(2))=(0.322mol)/((4.444+0.322)mol)=0.068`
`x_(H_(2)O)=1-0.068=0.92`