Question

6. A 0.004M solution of Na2SO4 is
isotonic with a 0.01M solution of glucose
at the 25°C. The degree of dissociation is
(A) 25% (B) 50% (C) 75% (D) 85%​

Answer 1

$\huge\underline\mathbb\pink{ANSWER\::}$

(C) 75%

Given, 0.004 M solution of Na2SO4 is isotonic with 0.01 M solution of glucose.

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

II Na2 SO4 = Il glucose = MRT

Or,

ix 0.004RT = 0.01RT

i= 2.5

Now, Dissociation of Na SO, is as follows:

Na2SO4 = ${2Na}^{+}$ + ${SO4}^{2-}$

1- a 2a a

So,

i = 1- a+ 2a + a

i = 1+ 2a

=> 2.5 = 1+ 2a

=> a = 0.75

The degree of dissociation = $\frac{75}{100}$×100%

The degree of dissociation = 75%

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