"If A2B is 30% ionised in an aqueous solution, then the value of van't Hoff factor (i) is _______ × 10−1."

Question

Accepted Answer

"

Percent ionisation of $A_2B=30\%$

The dissociation of $A_2B$ in aqueous solution can be represented as

$A_2B \rightarrow 2A^+ + B^{2-}$

1 mole of $A_2B$ 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 $A_2B$ 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,

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

$= 1 + 2 \times 0.3$

$= 1 + 0.6$

$= 1.6$

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

"

If A₂B is 30% ionised in an aqueous solution, then the value of van't Hoff factor (i) is _______ × 10⁻¹.