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0.004M Na₂SO₄ is isotonic with 0.01 M glucose. Degree of ionization of Na₂SO₄ is

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75%

25%

50%

85%

answer is C.

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Detailed Solution

$\begin{array}{l} π_{1} = π_{2}; {iCC}_{1} T = C_{2} RT \\ \Rightarrow i \times 0 .004 = 0 .01 \Rightarrow i = \frac{0 .01}{0 .004} = 2 .5 \\ α = \frac{i - 1}{n - 1} = \frac{1 .5}{2} = 0 .75 (or) 75 % \end{array}$

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