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Question: What will be the degree of dissociation of \(0.1\)M \(Mg(NO_{3})_{2}\) solution if van’t Hoff facto...

What will be the degree of dissociation of 0.10.1M

Mg(NO3)2Mg(NO_{3})_{2} solution if van’t Hoff factor is 2.742.74

A

75%

B

87%

C

100%

D

92%

Answer

87%

Explanation

Solution

Mg(NO3)2→Mg2++2NO3−Mg(NO_{3})_{2} \rightarrow Mg^{2 +} + 2NO_{3}^{-}

α=i−1n−1=2.74−13−1=1.742=0.87\alpha = \frac{i - 1}{n - 1} = \frac{2.74 - 1}{3 - 1} = \frac{1.74}{2} = 0.87

Degree of dissociation =0.87×100=87%= 0.87 \times 100 = 87\%