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The Van’t Hoff factor for $0.1 M Al {({NO}_{3})}_{3}$ solution is 2.5, then the percent dissociation of aluminium nitrate is

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74%

81%

50%

100%

answer is A.

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Detailed Solution

$A l {(N O_{3})}_{3} \to \overset{3 +}{A l} + 3 N O_{3}^{-} (n = 4)$

$α = \frac{i - 1}{n - 1} = \frac{2.5 - 1}{4 - 1} = \frac{1.5}{3} = 0.5$

$% α = α \times 100 = 0.5 \times 100 = 50 %$

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