`1.0 g`of a non-electrolyte solute (molar mass`250gmol^(-1)` was dissolved in 51.2 g of benzene. If the freezing point depression constant of benzene is `5.12K kgmol^(-1)` the lowering in freezing point will be

Options

A. 0.4 K
B. 0.3 K
C. 0.5 K
D. 0.2 K

Answer

A

Explanation

Molality of non-electrolyte solute
`=(("Weight of solute in (g)")/("molecular weight of solute"))/("weight of solvent in (kg)")=((1)/(250))/(0.0512)`
`=(1)/(250xx0.0512)=0.0781 m`
`Delta T_(f)=K_(f)xx` molality of solution
`=5.12xx0.0781 ~~ 0.4 K`