Question

0.004 M $N{a}_{2}S{O}_4$ is isotonic with 0.01 M Glucose. Degree of dissociation of $N{a}_{2}S{O}_4$ is:

• A. 75%
• B. 50%
• C. 25%
• D. 85%

Answer 1

The correct option is A 75%

Given, 0.004 M solution of $N{a}_{2}S{O}_4$ is isotonic with 0.01 M solution of glucose.

Therefore, According to vant Hoff,s ideal gas equation,

${\Pi }_{N{a}_{2}S{O}_4}={\Pi }_{glucose}=MRT$

Or,

$i\times 0.004RT=0.01RT$

$\Rightarrow i=2.5$

Now, Dissociation of $N{a}_{2}S{O}_4$ is as follows:

$$\begin{array}{ccccc}N{a}_{2}S{O}_4& \rightleftharpoons & 2N{a}^{+}& +& S{O}_4^{2-}\\ 1-\alpha & & 2\alpha & & \alpha \end{array}$$

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%