Question

$\text{A 0.004 M}N{a}_{2}S{O}_4$ aqueous solution is isotonic with $0.01\text{M}$ glucose at $300\text{K}$. Thus, the degree of dissociation of $N{a}_{2}S{O}_4$ is

• A. 50%
• B. 75%
• C. 25%
• D. 85%

Answer 1

The correct option is B 75%

$N{a}_{2}S{O}_4\to 2N{a}^{+}+S{O}_4^{2-}$
Hence, n = 3 , $\alpha =1$
$i=1+(n-1)\alpha$
$i=1+2\alpha$
${\pi }_1=CRTi$
${\pi }_1=0.004\times R\times 300\times (1+2\alpha )$
As glucose is a non - electrolyte, i = 1
${\pi }_2=CRTi$
${\pi }_2=0.01\times R\times 300\times 1$
Two solution are isotonic
$0.004\times R\times 300\times (1+2\alpha )$ = $0.01\times R\times 300\times 1$
$(1+2\alpha )$ = $\frac{0.01}{0.004}$
$\alpha$ = 0.75
thus, dissociation = 75%