What will be the degree of dissociation of $0.1 M$ $M g {({N O}_{3})}_{2}$ solution if van't Hoff factor is $2.74$ ?

75

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87

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100

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92

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Solution

The correct option is A $87$ %
The van't Hoff factor is the ratio between the actual concentration of particles produced when the substance is dissolved and the concentration of a substance as calculated from its mass. For most non-electrolytes dissolved in water, the van't Hoff factor is essentially 1.

$M g (N O_{3})_{2} \to M g^{2} + + 2 N O_{3}^{-}$
$α = \frac{i - 1}{n - 1}$
= $\frac{2.74 - 1}{3 - 1}$
= $\frac{1.74}{2}$
= $0.87$
Degree of Dissociation = $0.87 (100) = 87 %$

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