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0.002 $M {NaSO}_{4}$ is isotonic with 0.005 M glucose solution. The degree of dissociation of ${Na}_{2} {SO}_{4}$ is

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75%

95%

50%

35%

answer is B.

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Detailed Solution

For isotonic solution
$i_{1} C_{1} = i_{2} C_{2}$

For glucose $i = 1, {Na}_{2} {SO}_{4} = 3$
$0 .005 = i_{2} \times 0 .002$
$i_{2} = 2 .5 ∴ α = \frac{i - 1}{n - 1}$
$α = \frac{2.5 - 1}{3 - 1} = \frac{1.5}{2} = 0.75 or 75 %$

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