SolutionHard

Question

A solution of ‘x’ mole of sucrose in 100 g of water freezes at − 0.2°C. As ice separates out, the freezing point goes down to − 0.25°C. How many grams of ice would have separated?

Options

A.18 g
B.20 g
C.80 g
D.25 g

Solution

$\Delta T_{f} = K_{f}.m$

$\therefore 0.2 = K_{f} \times \frac{x}{100/1000}\text{ and 0.25 = }K_{f} \times \frac{x}{y/1000}$

∴ Mass of ice separated out = 100 – y = 20 gm

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