JEE Main 2025 — Solutions Question with Solution

Percent ionisation of $A_{2}B=30\%$

The dissociation of $A_{2}B$ in aqueous solution can be represented as

$A_{2}B\to 2A^{+}+B^{2-}$

1 mole of $A_{2}B$ produces 2 moles of $A^{+}$ and 1 mole of $B^{2-}$ ions.

The total number of moles of ions produced (n) from the dissociation is

$n=2+1=3$

The degree of dissociation given that $A_{2}B$ is 30% ionised, the degree of dissociation $(\lambda )$ or degree of ionisation

$\lambda =\frac{30}{100}=0.3$

Van't Hoff factor ($i$) can be calculated using the formula,

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

Substitute the values of $n$ and $\lambda$ as,

$$i=1+(3-1)\times 0.3$$

$$=1+2\times 0.3$$

$$=1+0.6$$

$$=1.6$$

$$=16\times 10^{-1}$$