In what volume ratio can 0.1M ${Fe}_{2} ({SO}_{4})_{3}$ and 0.1M ${Al}_{2} ({SO}_{4})_{3}$ be mixed so that the ratio of total number of moles of positive ions to that of negative ions in the solution is 2:3? Consider ${Fe}^{3 +}, {Al}^{3 +}$ and ${SO}_{4}^{2 -}$ as only ions in the solution. Salts are completely dissociated and do not undergo any hydrolysis.

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$\begin{aligned} 1 : 1 \end{aligned}$

$3 : 16$

Any of these values

$4 : 7$

answer is D.

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Detailed Solution

$\frac{V_{1} \times 0.1 \times 2 + V_{2} \times 0.1 \times 2}{V_{1} \times 0.1 \times 3 + V_{2} \times 0.1 \times 3} = \frac{2}{3}$

So, any value of $V_{1}$ and $V_{2}$ are possible.

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