Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  AIMS  >>  Class12  >>  Chemistry  >>  Liquid Solutions
0 votes

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

Can you answer this question?

1 Answer

0 votes
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

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App