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 $\left(K_f=1.86Kkgmo{l}^{-1}\right)$

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

Answer 1

Answer: (D)

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

$=400g$