Question

The van't Hoff factor (i) for a dilute aqueous solution of the strong electrolyte barium hydroxide is:

• A. 2
• B. 3
• C. 0
• D. 1

Answer 1

The correct option is B 3

For strong electrolytes, Van't Hoff factor (i) to the number of ions produced.
$\alpha =\frac{i-1}{n-1}$
Here, $\alpha$ is the degree of dissociation,
i = van't Hoff factor
n = number of ions
$Ba(OH{)}_2$ is a strong electrolyte that dissociates 100% isn an aqueous medium as:
$Ba(OH{)}_2\to B{a}^{2+}(aq)+2O{H}^{-}(aq)$
So, number of ions n = 1 + 2 = 3
As barium hydroxide is a strong electrolyte,
$\therefore \alpha =1$
Using formula,
$\alpha =\frac{i-1}{n-1}$
$\Rightarrow 1=\frac{i-1}{3-1}$
$\Rightarrow (i-1)=2$
$\Rightarrow i=2+1=3$