JEE Chemistry 2025 Question with Solution

Given: $\text{A}_2\text{B}$ is $30\%$ ionised in aqueous solution.

Find: The value of van't Hoff factor $i$.

For $1$ mole of $\text{A}_2\text{B}$, the dissociation is

$$\text{A}_2\text{B} \rightarrow 2\text{A}^+ + \text{B}^-$$

So, on complete ionisation, $1$ mole of $\text{A}_2\text{B}$ gives a total of $3$ moles of ions.

Since the compound is only $30\%$ ionised, moles ionised $= 0.3$ and moles undissociated $= 0.7$.

From $0.3$ moles of ionised $\text{A}_2\text{B}$:

• moles of $\text{A}^+ = 2 \times 0.3 = 0.6$
• moles of $\text{B}^- = 0.3$

Therefore, total moles after ionisation are

$$0.7 + 0.6 + 0.3 = 1.6$$

Now,

$$i = \frac{\text{total moles after ionisation}}{\text{initial moles of solute}} = \frac{1.6}{1} = 1.6$$

Therefore, the van't Hoff factor is $1.6$.

Given: Degree of ionisation $\alpha = 0.3$ for $\text{A}_2\text{B}$.

Find: van't Hoff factor $i$.

For a solute that dissociates into $n$ particles, the relation is

$$i = 1 + (n-1)\alpha$$

Here, $\text{A}_2\text{B}$ dissociates into $2\text{A}^+ + \text{B}^-$, so total particles after dissociation $n = 3$.

Substituting,

$$i = 1 + (3-1)(0.3)$$ $$i = 1 + 0.6 = 1.6$$

Therefore, the van't Hoff factor is $1.6$.