Question

A $5\%$ solution of glucose is isotonic with $1.1\%$ solution of KCl at ${30}^{o}C$. Calculate the degree of ionisation of KCl.

• A. $90\%$
• B. $88\%$
• C. $78\%$
• D. $80\%$

Answer 1

The correct option is B $88\%$

Let's assume the mass of the solution = 100 g

so, mass of glucose = 5 g

and mass of KCl = 1.1 g

For isotonic solution-

${\pi }_1={\pi }_2⟶$ $C_{1}RT=i{C}_{2}RT$

$C_1=i{C}_2$

$\frac{5}{180}=i\frac{1.1}{74.5}$

$i=1.88$

we have,

Degree of ionisation, $\alpha =\frac{i-1}{n-1}$

For,

$K⟶K^{+}+C{l}^{-1}$

Thus, n=2

$\alpha =\frac{1.88-1}{2-1}=0.88$

Degree of ionization, $\alpha$ = $0.88\times 100=88$%