Question

An aqueous solution containing $3g$ of a solute of molar mass $111.6gmo{l}^{-1}$ in a certain mass of water freezes at $-{0.125}^{\circ }C$. The mass of water in grams present in the solution is $(K_f=1.86Kkgmo{l}^{-1})$.

• A. $300$
• B. $600$
• C. $500$
• D. $400$
• E. $250$

Answer 1

The correct option is D $400$

Water freezes at $-{0.125}^{\circ }C$
The freezing point of water $=0^{\circ }C$

Depression in freezing point,
$△T_f=0-(-0.125)$
$=+0.125$

We know that
$△T_f=\frac{K_f\times w\times 1000}{M\times W}$

(where, $w$ and $M=$ weight and molar mass of solute and $W=$ weight of solvent or water)

On substituting values, we get

$0.125=\frac{1.86\times 3\times 1000}{111.6\times W}$

or $W=\frac{1.86\times 3\times 1000}{0.125\times 111.6}$

$W=400g$.