Pure benzenen boils at `180^(@)C`. If latent heat of vaporization of benzenen is `90 "cal"` per `g`, calculate the molecular weight of solute.

Answer

`m=189.79`

Explanation

B.pt. of benzene `=80+273=353K`
Latent heat of vaporization for `C_(6)H_(6)=90 "cal"//g`
`therefore K_(b)=(RT^(2))/(1000l)=(2xx(353)^(2))/(1000xx90)=2.77K "mol"^(-1)kg`
`because DeltaT=80.175-80=0.175, w=83.4g`
`because DeltaT=(100lambdak_(b)w)/(mxxW)`
`0.175=(1000xx2.77xx1)/(mxx83.4)`
`therefore m=189.79`