Question

A 5% solution of glucose is isotonic with 1.1 % solution of KCl at ${30^o}C$. Calculate the degree of ionization of KCl.
A. 90%
B. 88%
C. 78%
D. 80%

Answer 1

Hint: There is a relationship between degree of ionization and Van’t Hoff factor and it is as follows.
\[\alpha =\dfrac{i-1}{n-1}\]
Where $\alpha $ = degree of ionization
i = Van’t Hoff factor
n = number of ions produced when solute is dissociated.

Complete Solution :
- In the question it is given that the 5% solution of glucose is isotonic with 1.1 % solution of KCl.
- We have to calculate the degree of ionization from the given data.
- Assume that the mass of the total solution is 100 g.
- Then the mass of the glucose will be 5 g and the mass of the KCl is 1.1 g.
- The molecular weight of the glucose is 180 and the molecular weight of the KCl is 74.5.
- We know that for isotonic solution
\[\begin{align}
 & {{\pi }_{1}} = {{\pi }_{2}} \\
 & {{C}_{1}}RT = i{{C}_{2}}RT \\
 & {{C}_{1}} = i{{C}_{2}} \\
\end{align}\]
Here ${{C}_{1}}$ = number of moles of glucose
${{C}_{2}}$ = number of moles of KCl

- Substitute all the known values in the above formula to get the Van’t Hoff factor value ‘I’.
\[\begin{align}
 & {{C}_{1}} = i{{C}_{2}} \\
 & \dfrac{5}{180} = i\dfrac{1.1}{74.5} \\
 & i=1.88 \\
\end{align}\]
- Then Van’t Hoff factor i = 1.88.

- Substitute the ‘i' value and ‘n’ value in the formula to get the degree of ionization of KCl.
\[\begin{align}
 & \alpha =\dfrac{i-1}{n-1} \\
 & \alpha =\dfrac{1.88-1}{2-1} \\
 & \alpha = 0.88 \\
\end{align}\]
Here n = number of ions produced when KCl is ionized in the solution. ($KCl\to {{K}^{+}}+C{{l}^{-}}$ )
- The degree of ionization = $0.88\times 100 = 88\text{ % }\!\!\!\!\text{ }$
- Therefore the degree of ionization of KCl is 88%.
So, the correct answer is “Option B”.

Note: To calculate the degree of ionization of KCl we need the value of ‘i'. To get the value of ‘i’ we should use the given data in the question to calculate it by taking the molecular weights of the solutes into consideration.