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(B) We know that $\Delta T_f = K_f \cdot m$. Since the amount of solute is constant, $\Delta T_f \propto \frac{1}{W}$, where $W$ is the mass of the so

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$20$

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(B) We know that $\Delta T_f = K_f \cdot m$. Since the amount of solute is constant, $\Delta T_f \propto \frac{1}{W}$, where $W$ is the mass of the solvent. Given: $\Delta T_{f1} = 0.2^\circ	ext{C}$, $W_1 = 100 	ext{ g}$, $\Delta T_{f2} = 0.25^\circ	ext{C}$. $\frac{\Delta T_{f1}}{\Delta T_{f2}} = \frac{W_2}{W_1}$ $\frac{0.2}{0.25} = \frac{W_2}{100}$ $W_2 = \frac{0.2 \cdot 100}{0.25} = 80 	ext{ g}$ Mass of ice separated = $W_1 - W_2 = 100 	ext{ g} - 80 	ext{ g} = 20 	ext{ g}$.

$A$ fixed amount of glucose is dissolved in $100 \text{ g}$ water to form a solution that freezes at $-0.2^\circ\text{C}$. If the solution is cooled down to $-0.25^\circ\text{C}$, then ......... $\text{g}$ of ice would have separated.

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