When $0.004 M N a_{2} S O_{4}$ is isotonic with 0.01M solution of glucose at same temperature.the apparent degree of dissociation of sodium sulphate 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%

For $N a_{2} S O_{4}, π_{1} = i M R T$
$= i \times 0.004 \times R T$
For glucose, $π_{2} = C R T$
= 0.01 RT
$π_{1} = π_{2}$ (isotonic)
or 0.004i = 0.01
i = 2.5
$\begin{matrix} N a_{2} S O_{4} ⇌ & 1 N a^{\oplus} + & S O_{4}^{2 -} 1 & 0 & 0 1 - α & 2 α & α \end{matrix}$
Total particles= $1 - α + 2 α + α$
$= 1 + 2 α ∴ i = \frac{1 + 2 α}{1} = 2.5$
$α = 0.75$ or 75% dissociated

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