KEAM 2014 Physics and Chemistry Paper

Water freezes at =0.125∘C.
Freezing point of water =0∘C
Depression in freezing point,
∆Tf=0−(−0.125)
=+0.125
We know that
∆Tf=‌

Kf×w×1000

M×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=‌

1.86×3×1000

111.6×W

or W=‌

1.86×3×1000

0.125×111.6

=400g