For 100 g of the solution
Mass of benzene = 30 g
Mass of carbon tetrachloride `=100-30=70"g"`
Molar mass of benzene `(C_(6)H_(6))=(12xx6)+(6xx1)=72+6=78"g mol"^(-1)`
Moles of benzene, `n_(C_(6)H_(6))=("Mass")/("Molar mass")=(30)/(78)=0.385" mol"`
Molar mass of carbon tetrachloride `(C Cl_(4))=12+(35.5xx4)`
`=12+142.0=154" g mol"^(-1)`
Moles of `C Cl_(4),n_(C Cl_(4))=(70"g")/((154" mol"^(-1)))=0.454" mol"`
Mole fraction of benzene, `x_(C_(6)H_(6))=(n_(C_(6)H_(6)))/(n_(C_(6)H_(6))+n_(C Cl_(4)))=(0.385" mol")/((0.385+0.454)" mol")=0.459`