30% by mass of `C_(6)H_(6)` in `C Cl_(4) implies30gC_(6)H_(6)` in 100 g solution
`therefore` no. of moles of `C_(6)H_(6), (.^(n)C_(6)H_(6))=30//78=0.385`
(molar mass of `C_(6)H_(6)=78g`)
no. of moles of
`C Cl_(4)(.^(n)C Cl_(4))=(70)/(154)=0.455`
`x_(C_(6)H_(6))=(nC_(6)H_(6))/(n_(C_(6)H_(6))+n_(C C l_(4)))`
`=(0.385)/(0.385+0.455)=(0.385)/(0.84)=0.458`
`x_(C Cl_(4))=1-0.458=0.542`
