`1.0 g` of a non-electrolyte solute( mol. Mass `250.0 g mol^(-1)`) was dissolved in `5.12 g` benzene. If the freezing point depression constant, `K_(f)` of benzene is `51.2 K kg mol^(-1)`, the freezing point of benzene will be lowered by:

Options

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

Answer

2

Explanation

We need to calculate `DeltaT_(f)` (depression of freezing point of solvent) which is given as
`DeltaT_(f)=iK_(f)m=iK_(f) W_(solute)/(mm_(solute)W_(solv))`
we are given
`i=1` as the solute is nonelectrolyte
`K_(f)=5.12 K kg mol^(-1)`
`W_(solute)=1.00 g`
`W_(solvent)=51.2/1000kg`
`mm_(solute)=250 g mol^(-1)`
Substituting these results, we get
`DeltaT_(f)=(1)(5.12 K kg mol^(-1))((1.00 g))/((250 g mol^(-1))(5.12/1000kg))`
`=0.4 K`