$0.004 M N a_{2} S O_{4}$ is isotonic with $0.01 M$ glucose. Degree of dissociation of $N a_{2} S O_{4}$ is:

75 %

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50 %

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25 %

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85 %

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Solution

The correct option is A $75 %$
Since both $N a_{2} S O_{4}$ and glucose solutions are isotonic,
$π_{N a_{2} S O_{4}} = π_{G l u c o s e}$
$i \times 0.004 R T = 0.01 R T$
$i = 2.5$
$N a_{2} S O_{4} \to 2 N a^{+} + S O_{4}^{2 -}$
$1 - α 2 α α$
$i = 1 + 2 α = 2.5$
$α = 0.75 o r 75 %$

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The correct option is A $75 %$
Since both $N a_{2} S O_{4}$ and glucose solutions are isotonic,
$π_{N a_{2} S O_{4}} = π_{G l u c o s e}$
$i \times 0.004 R T = 0.01 R T$
$i = 2.5$
$N a_{2} S O_{4} \to 2 N a^{+} + S O_{4}^{2 -}$
$1 - α 2 α α$
$i = 1 + 2 α = 2.5$
$α = 0.75 o r 75 %$