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0.004 M $$\mathrm{N}\mathrm{a}_{2}\mathrm{S}\mathrm{O}_{4}$$ is isotonic with 0.01 M Glucose. Degree of dissociation of  $$\mathrm{N}\mathrm{a}_{2}\mathrm{S}\mathrm{O}_{4}$$  is:


A
75%
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B
50%
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C
25%
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D
85%
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Solution

The correct option is A 75%
Given, 0.004 M solution of $$Na_2SO_4$$ is isotonic with 0.01 M solution of glucose.

Therefore, According to vant Hoff,s ideal gas equation,
$$\Pi_{Na_2SO_4} = \Pi_{glucose} = MRT $$
Or, 
$$i \times 0.004RT = 0.01RT$$
$$\Rightarrow i=2.5$$

Now, Dissociation of $$Na_2SO_4$$ is as follows:
\begin{matrix} Na_{ 2 }SO_{ 4 } & \rightleftharpoons  & 2Na^{ + } & + & SO_{ 4 }^{ 2- } \\ 1-\alpha  &  &2\alpha  &  &  \alpha \end{matrix} 
So, 
$$i = 1-\alpha + 2\alpha+\alpha$$
$$i=1+2\alpha$$
$$\Rightarrow 2.5 = 1+2\alpha$$
$$\Rightarrow \alpha = 0.75$$
Hence, degree of dissociation = 75%

Chemistry

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