Addition of `0.643g` of a compound to `50mL` of benzene (density: `0.879g mL^(-1))`lower the freezing point from `5.51^(@)C` to `5.03^(@)C`. If `K_(f)` for benzene is `5.12 K kg mol^(-1)`, calculate the molar mass of the compound.

Explanation

The molar mass ,`M_(B)` is calculated as :
`M_(B) = (k_(f) xx w_(B) xx 1000)/(w_(A) xx Delta T_(f))`
`w_(B) = 0.643 g`
`w_(A) = (50 mL) xx (0.879 g//mL)`
` = 43.95 g`
`k_(f) = 5.12 K m^(-1)`
`DeltaT_(f) = 5.51 - 5.03 = 0.48^(@)C`
`M_(B) = (5.12 xx 0.643 xx 1000)/(43.95 xx 0.48)`
= 156.06