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?

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20 gram

25 gram

18 gram

23 gram

answer is B.

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Detailed Solution

${ΔT}_{f} = m \times K_{f}$

$\frac{∆ T_{f_{1}}}{∆ T_{f_{2}}} = \frac{m_{1}}{m_{2}} \frac{- 0.2}{- 0.25} = \frac{\frac{x \times 1000}{100}}{x \times \frac{1000}{W (In grams)}} \frac{0.2}{0.25} = \frac{W}{100}$

$W = 80 gr$

$Hence, the grams of ice that would be seperated = 100 - 80 = 20 gr$

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