Addition of `0.643`g of compound to `50`ml of benzene (density `=0.879mol^(-1)` ) lower the freezing point from`50.51^(@)C`. If `k_(f)` for benzene is `5.12` the molecule of the compound is

Options

A. `156.00`
B. `312.00`
C. `78.00`
D. `468.000`

Answer

A

Explanation

Weight of bezene =`Vxxd=50xx0.879g`
`W=43.95g`
Weight of compound =`0.643gm`
`DeltaT_(f)=50.51-50.03=0.48`
`:.DeltaT_(f)=K_(f)xx(w)/(m)xx(1000)/(w)`
`0.48=5.12xx(0.643)/(m)xx(1000)/(43.95)`
` :.m=156.05`