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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95 %

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Solution

The correct option is A $75 %$
$π_{N a_{2} S O_{4}} = π_{g l u c o s e}$
$i_{1} C_{1} R T = i_{2} C_{2} R T$
i for glucose = 1
$i \times 0.004 = 0.01$
$i = \frac{0.01}{0.004} = 2.5$
$\begin{matrix} N a_{2} S O_{4} ⇌ & 2 N a^{+} + & S O_{4}^{2 -} 1 & - & - 1 - α & 2 α & α \end{matrix}$
$i = 1 - α + 2 α + α = 1 + 2 α$
$2.5 = 1 + 2 α$
$α = 75 %$

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