Question

A sol contanng 1.9 g per 100 ml of kcl (m= 74.5g/mol) is isotonic with a sol . Containg 3g/100 ml of urea( m=60g/mol) calculate the degree of dissociation of kcl solution. Assume that both the sol have same temperature

Answer 1

Answer: 25%

Explanation: Isotonic solutions are those solutions which have the same osmotic pressure.  If osmotic pressures are equal at the same temperature, concentrations must also be equal.

$\pi =CRT$ for non electrolytes such as urea

$\pi =iCRT$ for electrolytes such as $KCl$

$\pi$ = osmotic pressure

C= concentration

R= solution constant

T= temperature

i= vant hoff factor

Thus: $C_{urea}=i\times C_{KCl}$ where concentration is in molarity.

For urea solution: 3 g of urea is dissolved in 100 ml of solution.

For $KCl$ : 1.9 g of $KCl$  is dissolved in 100 ml of solution.

$Molarity=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{volume of solution in ml}}$

$C_{urea}=\frac{3\times 1000}{60}\times {100}=0.5$

$C_ {KCl}=\frac{1.9\times 1000}{74.5}\times 100=0.25$

Thus: $C_{urea}=i\times C_{KCl}$ where concentration is in molarity.

$0.5M=i\times 0.25M$

$i=2.0$

$i=\frac{\text {observed colligative property}}{\text {Calculated Colligative property}}$

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

 0.25            0             0

$0.25-\alpha$        $\alpha$          $\alpha$  

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

thus $i=\frac{0.25+\alpha}{0.25}$

$2.0=\frac{0.25+\alpha}{0.25}$

$\alpha=0.25$

$\alpha=25\%$

Thus the degree of dissociation of KCl solution is 25%.