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Q.

0.004 M solution of Na₂SO₄ is isotonic with 0.01 M solution of glucose at same temperature. The approximate degree of dissociation of Na₂SO₄ is

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a

50%

b

25%

c

75%

d

85%

answer is C.

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Detailed Solution

$π_{N a_{2} S O_{4}} = π_{G l u \cos e}$
CSTi = CsTi
$4 \times 10^{- 3} \times i = 10^{- 2} \times 1 i_{N a_{2} S O_{4}} = \frac{10^{- 2}}{4 \times 10^{- 3}} = 2.5 α_{N a_{2} S O_{4}} = \frac{i - 1}{n - 1} = \frac{2.5 - 1}{3 - 1} = \frac{1.5}{2} = 0.75 α % = 75$

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