An aqueous solution containing 3 g of a solute of molar mass `111.6 g mol^(-1)` in a certain mass of water freezes at `-0.125^(@)C.` The mass of water in grams present in the solution is `(K_(f) = 1.86 K kg mol^(-1))`

Options

A. a) 300
B. b) 600
C. c) 500
D. d) 400

Answer

D

Explanation

Water freezes at `= 0.125 ^(@)C.`
Freezing point of water `= 0 ^(@)C`
Depression in freezing point,
`Delta T _(f) = 0 - (-0.125)`
`= + 0.125`
We know that
`Delta T _(f) = ( K _(f) xx w xx 1000)/( M xx W)`
(where, w and M= weight and molar mass of solute and W = weight of solvent or water)
On substituting values, we get
`0.125 = (1. 86 xx 3 xx 1000)/( 111.6 xx W)`
or `W = (1. 86 xx 3 xx 1000)/(0.125 xx 111.6)= 400 g`