Calculate the mole fraction of benzene in solution containing 30% by mass in carbontetrachloride.

Explanation

Let the total mass of the solution be 100 g and the mass of benzene be 30 g.
`therefore` Mass of carbon tetrachloride `= (100-30)g=70g`
Molar mass of benzene `(C_(6)H_(6))=(6xx12+6xx1)g mol^(-1)`
`= 78 mol^(-1)`
`therefore` Number of moles of `= C_(6)H_(6)`
`=(30)/(78)mol = 0.3846 mol`
Molar mass of carbon tetrachloride `(C Cl_(4))`
`= (1xx12)+(4xx35.5)=154`
`therefore` Number of moles of `C Cl_(4)=(70)/(154)=0.4545` mol
Thus, the mole fraction of `C_(6)H_(6)` is given as :
`= ("Number of moles of "C_(6)H_(6))/("Number of moles of "C_(6)H_(6) + "Number of moles " C Cl_(4))`
`=(0.3846)/(0.3846+0.4545)=0.458`