Let mass of solution =100 g
∴‌‌ Mass of benzene in the solution =30 g
∴‌‌ Mass of carbon tetrachloride =100−30 g=70g
Molar mass of benzene ( C6H6)=78gmol−1
∴‌ No. of moles of benzene =

Mass

‌Molar‌mass

=

30g

78gmol−1

=0.385
∴‌ No. of moles of benzene =

Mass

‌Molar‌mass

=

70g

154gmol−1

=0.385
Total number of moles of solution =0.385+0.455=0.84
Mole fraction of benzene =

‌Moles‌of‌benzene

Total‌moles‌in‌the‌solution

=

0.385

0.84

=0.458
∴ Mole fraction of CCl4=1–0.458=0.542