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If $α$ is the degree of dissociation of $N a_{2} S O_{4}$ , the van't Hoff factor used tor the calculation of molecular mass is

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detailed solution

Correct option is A

${Na}_{2} {SO}_{4}$ ionizes as

$\underset{n (1 - α)}{{Na}_{2} {SO}_{4} \to} \underset{n (2 α)}{2 {Na}^{+}} + \underset{4 α}{{SO}_{4}^{2 -}}$

Total amount of species in solution will be $n^{'} = n (1 - a) + 2 n α + n α = n (1 + 2 α)$

The van't Hoff factor is $i = \frac{n^{'}}{n} = 1 + 2 α$

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If α is the degree of dissociation of Na2SO4 , the van't Hoff factor used tor the calculation of molecular mass is