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Home  >>  AIMS  >>  Class12  >>  Chemistry  >>  Liquid Solutions
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The freezing point of aqueous solution contains 5% by mass urea, 1.0% by mass KCl and 10% by mass of glucose is: ($K_{H_2O} = 1.86 K molality^{–1}$)

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5 % urea by mass $\Rightarrow$ 5 g of urea is present in 100 g of solution or 95 g of water
1.0% KCl by mass $\Rightarrow$ 1 g of KCl is present in 100 g of solution or 99 g of water
10% glucose by mass $\Rightarrow$ 10 g of glucose is present in 100 g of solution or 90 g of water
Molar mass of urea = 60 g $mol^{−1}$
Molar mass of KCl = 74.5 g$ mol^{−1}$
Molar mass of urea = 180 g $mol^{−1}$
$\Delta T_f=K_fm=\large\frac{K_f\times W_2\times 1000}{W_1\times M_2}$
$\Delta T_f = \Delta T_f$ for urea + $\Delta T_f$ for KCl + ΔTf for glucose
$\Rightarrow \large\frac{1000\times 1.86\times 5}{95\times 60}+\frac{1000\times 1.86\times 1}{74.5\times 99}+\frac{1000\times 1.86\times 10}{90\times 180}$$=3.03^{\large\circ}$
Freezing point = 273 – 3.03 = 269 .97 K
Hence (A) is the correct answer.
answered Jun 12, 2014 by sreemathi.v
 

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