Question

1.9g of 100 ml solution of kcl is isotonic with 3g of 100 ml solution of urea .find degree of dissociation of kcl . assume same temperature for both.​

Answer 1

Answer: 100 %

Explanation: Isotonic solutions are those solutions which have the same osmotic pressure.

$\pi =CRT$ if osmotic pressures are equal at the same temperature, concentrations must also be equal.

$\pi$ = osmotic pressure

C= concentration

R= solution constant

T= temperature

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

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

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

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

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

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

As $C_{urea}=i\times C_{KCl}$

$0.5=i\times 0.25$

$i= 2$

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

Formula used :

$\alpha=\frac{i-1}{n-1}$

where,

n = number of particles after dissociation = 2

$\alpha=\frac{2-1}{2-1}=1$

Now we have to calculate the percentage dissociation.

$\text{percentage dissociation}=1\times 100=100$

Therefore, the percentage dissociation of $KCl$ is 100%.