Question

A solution of x moles of sucrose in 100 grams of water freezes at $-{0.2}^{o}C$. As ice separates the freezing point goes down to ${0.25}^{o}C$. How many grams of ice would have separated?

• A. $18$ grams
• B. $20$ grams
• C. $25$ grams
• D. $23$ grams

Answer 1

The correct option is B $20$ grams

The expression for the depression in the freezing point and the molality is

$K_f=\frac{\Delta T_f}{m}$

For a solution which freezes at $-{0.2}^{o}C,(\text{molality}{)}_f=\frac{x\times 1000}{100}=\frac{0.2}{K_f}$

For a solution which freezes ar $-{0.25}^{o}C,(\text{molality}{)}_f=\frac{x\times 1000}{w}=\frac{0.25}{K_f}$

On dividing equation (1) by equation (2), we get

$\frac{0.2}{0.25}=\frac{w}{100}$

Hence, $w=80g$

The grams of ice that would have separated $100-80=20g$

Hence, the correct option is $B$