Correct Answer - `0.458,0542`
`W_(C_(6)H_(6))=30g,W_(sol)=100g,W_( C Cl_(4))=(100-30)g`
`=70g`
`Mw` of `C_(6)H_(6)=78g mol^(-1),Mw of C Cl_(4)=12+4xx35.5`
`=154 g mol^(-1)`
`n_(C_(6)H_(6))=(W_(C_(6)H_(6)))/(Mw of c_(6)H_(6))=(30g)/(78g mol^(-1))=0.385`
`n_(C C l_(4))=(W_(C Cl_(4)))/(Mw of C C l_(4))=(70g)/(154g mol^(-1))=0.425`
`CHMi_(C_(6)H_(6))=(n_(C_(6)H_(6)))/(n_(C_(6)H_(6))+nC Cl_(4))=(0.385)/(0.385+0.425)=(0.385)/(0.84)=0.485`
`CHMi_(C C l_(4))=1-0.485=0.542`
