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.

Answer

`156.06 g//mol`

Explanation

Given
`DeltaT = (1000 xx K'_(f)xxw)/(m xxW)`
`W =` wt.of benzene `= V xx d = 50 xx 0.879g`
`:. 0.48 =(1000 xx 5.12 xx 0.643)/(m xx 50 xx 0.879)`
`DeltaT = 5.52 - 5.03 = 0.48`
`:. M = 156.06`
`w = 0.643g K'_(f) = 5.12 K mol^(-1) kg`