Question

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 $-0.125^{\circ} C$. The mass of water in grams present in the solution is $\left(K_{f}=1.86\, Kkg \,mol ^{-1}\right)$

• A. 300
• B. 600
• C. 500
• D. 400

Answer 1

Answer: (D)

400

Answer 2

Water freezes at $=0.125^{\circ} C$.

Freezing point of water $=0^{\circ} C$

Depression in freezing point,

$\Delta T_{f} =0-(-0.125)$
$=+0.125$

We know that

$\Delta T_{f}=\frac{K_{f} \times w \times 1000}{M \times W}$

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

On substituting values, we get

$125 =\frac{1.86 \times 3 \times 1000}{111.6 \times W}$

or $W =\frac{1.86 \times 3 \times 1000}{0.125 \times 111.6}$
$=400\, g$