Calculate the mass percentage of benzene `(C_(6)H_(6))` and carbon tetraCHMloride `( C Cl_(4))` if `22g` of benzene is dissoved in `122g` of carbon tetraCHMloride.

Answer

`C_(6)H_(6)=15.28%,C C l_(4)=84.72%`

Explanation

`W_(sol)=W_(C_(6)H_(6))+W_(C C l_(4))=(22+122)=144g`
Mass percentage of `C C l_(4) =(W_(C C l_(4)))/(W_(sol))xx100=(122g)/(144g)xx100`
Mass percentage of `C_(6)H_(6)=(W_(C_(6)H_(6)))/(W_("soln"))xx100=(122)/(144g)xx100`
`=15.28%`
Alternatively
Mass percentage of `C_(6)H_(6)=100-84.72=15.28%`