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The freezing point of aqueous solution that contains 5% by mass urea, 1.0% by mass KCl and 10% by mass of glucose is
$(K_{f} (H_{2} O) = 1.86 k m o l a l i t y^{- 1})$

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Detailed Solution

$T_{f}^{o} - T_{f} = K_{f} \times m$

$n_{1} = \frac{5}{60} n_{2} = \frac{1}{74.5}, n_{3} = \frac{10}{180}$

For KCl i = 2 then $n_{2} = 2 \times \frac{1}{74.5}$

$n = \frac{5}{60} + \frac{2}{74.5} + \frac{10}{180}$

$T_{f}^{o} - T_{f} = 1.86 \times 0.164 \times \frac{1000}{300}$

$273 - T_{f} = 1.0168$

$∴ T_{f} = 271.98$

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