A solution containing 1.9g\: \: per\: \: 100mL\: \: of\: \: KCl\left ( M=74.5gmol^{-1} \right ) is isotonic with a solution containing 3g\: per\: 100mL\: \: of\: \: urea\left ( M=60gmol^{-1} \right ). Calculate the degree of dissociation of KCl solution. Assume that both the solutions have same temperature. 

 

 

 

 

 
 
 
 
 

Answers (1)
S Sumit

For isotonic solutions,

C_{KCl}RT=C_{urea}

\Rightarrow i\times C_{KCl}=C_{urea}

\Rightarrow i\times\frac{1.9\times 1000}{74.5\times 100}=\frac{3}{60}\times\frac{1000}{100}

\Rightarrow i\times 0.25=0.5

\Rightarrow i=2

i=\frac{observed\: colligative\:property}{calculated\:colligative\:property}

KCl\rightarrow \: \: \: \: \: K^+\: \: +\: \: \: Cl^-

0.25                0               0

0.25-\alpha       \alpha              \alpha

Total moles after dissociation = 0.25-\alpha+\alpha +\alpha

                                              =0.25+\alpha

i=\frac{0.25+\alpha}{0.25}\Rightarrow 2=\frac{0.25+\alpha}{0.25}\Rightarrow \alpha=0.25

Related Chapters

Preparation Products

Knockout NEET May 2021 (One Month)

An exhaustive E-learning program for the complete preparation of NEET..

₹ 14000/- ₹ 6999/-
Buy Now
Foundation 2021 Class 10th Maths

Master Maths with "Foundation course for class 10th" -AI Enabled Personalized Coaching -200+ Video lectures -Chapter-wise tests.

₹ 350/- ₹ 112/-
Buy Now
Knockout JEE Main April 2021 (One Month)

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 14000/- ₹ 6999/-
Buy Now
Knockout NEET May 2021

An exhaustive E-learning program for the complete preparation of NEET..

₹ 22999/- ₹ 14999/-
Buy Now
Knockout JEE Main April 2021

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 22999/- ₹ 14999/-
Buy Now
Boost your Preparation for JEE Main with our Foundation Course
 
Exams
Articles
Questions