If $$A_{2}B$$ is 30% ionised in an aqueous solution, then the value of van't Hoff factor (i) is ______ $$\times 10^{-1}$$

If $$A_2B$$ is 30% ionised in an aqueous solution, find the van't Hoff factor (i) expressed as $$\_\_\_ \times 10^{-1}$$.

The dissociation equation is $$A_2B \rightarrow 2A^+ + B^{2-}$$, and one formula unit produces 3 ions (2 cations + 1 anion), so $$n = 3$$.

The van't Hoff factor formula is $$i = 1 + \alpha(n - 1)$$, where $$\alpha = 0.30$$ and $$n = 3$$. Thus, $$i = 1 + 0.30(3 - 1) = 1 + 0.60 = 1.60$$.

Expressing this in the required form gives $$i = 1.60 = 16 \times 10^{-1}$$, so the answer is 16.