Addition of 0.643g of a compound to 43.95g of benzene lowers the freezing point from `5.51^@`C to `5.03^@`C. If `K_f` for benzene is 5.12K kg `mol^-1` , calculate the molar mass of the compound.

Explanation

Mass of solute = 0.643 g
Volume of benzene = 50 mL
Density = 0.879 g `mL^(-1)`
Mass of benzene = 43.95 g
`K_(f)=5.12" K kg mol"^(-1)`
`DeltaT_(f)=T_(f)^(@)-T_(f)=5.51-5.03=0.48^(@)C`
`M_(B)=(K_(f)xxW_(B)xx1000)/(DeltaT_(f)xx"Mass of solvent (g)")=(5.12xx0.643xx1000)/(0.48xx43.95)=156"mol"^(-1)`