Aqueous solution of 0.004 M $N a_{2} S O_{4}$ and 0.01 M glucose are isotonic. The degree of dissociation of $N a_{2} S O_{4}$ is

25%

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60%

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75%

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85%

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Solution

The correct option is C 75%
$N a_{2} S O_{4} \to 2 N a^{+} + S O_{4}^{- 2}$
$0.004 (1 - α) 0.008 α 0.004 α$
$N o w$
$0.004 (1 - α) + 0.008 α + 0.004 α = 0.01$
$0.008 α = 0.01 - 0.004$
$0.008 α = 0.006$
$α = \frac{6}{8} = 0.75$
$P e r c e n t a g e d i s s o c i a t i o n = 75 %$

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Aqueous solution of 0.004 M Na2SO4 and 0.01 M glucose are isotonic. The degree of dissociation of Na2SO4 is

The correct option is C 75%Na2SO4→2Na++SO−24 0.004(1−α)0.008α0.004α Now 0.004(1−α)+0.008α+0.004α=0.01 0.008α=0.01−0.004 0.008α=0.006 α=68=0.75 Percentage dissociation=75%

Aqueous solution of 0.004 M $N a_{2} S O_{4}$ and 0.01 M glucose are isotonic. The degree of dissociation of $N a_{2} S O_{4}$ is

The correct option is C 75%
$N a_{2} S O_{4} \to 2 N a^{+} + S O_{4}^{- 2}$
$0.004 (1 - α) 0.008 α 0.004 α$
$N o w$
$0.004 (1 - α) + 0.008 α + 0.004 α = 0.01$
$0.008 α = 0.01 - 0.004$
$0.008 α = 0.006$
$α = \frac{6}{8} = 0.75$
$P e r c e n t a g e d i s s o c i a t i o n = 75 %$