Dissolution of 1.5 g of a non-volatile solute `(mol.wt.=60)` in 250 g of a solvent reduces its freezing point by `0.01^(@)C`. Find the molal depression constant of the solvent.

Options

A. `0.01`
B. `0.001`
C. `0.0001`
D. `0.1`

Answer

D

Explanation

Depression in freezing point,
`DeltaT_(f)=K_(f)xxm`
`"Molality"=("Weight of solute"xx1000)/("Molecular weight of solute "xx" weightof solvent")`
`=(1.5xx1000)/(60xx250)=0.1"molar"`
`rArrDeltaT_(f)=K_(f)xx0.1`
`rArr0.01=K_(f)xx0.1`
`:.K_(f)=(0.01)/(0.1)=0.1`