Question

A $0.004M$ solution of ${Na}_{2}S{O}_4$ is isotonic with a $0.010M$ solution of glucose at the $25$. The apparent degree of dissociation of ${Na}_{2}S{O}_4$ is:

• A. $25$%
• B. $50$%
• C. $75$%
• D. $85$%

Answer 1

The correct option is C $75$%

Since two solutions are isotonic, the product of vant Hoff factor i and the concentration C is same. $i_{Na2SO4}\times C_{Na2SO4}=iglucose\times C_{glucose}$
$i_{Na2SO4}\times 0.004M=1\times 0.010M$
$i_{Na2SO4}=2.5$
One molecule of $N{a}_{2}S{O}_4$ on dissociation gives three ions
$N{a}_{2}S{O}_4\to 2N{a}^{+}+S{O}_4^{2-}$
$n=3$
The apparent degree of dissociation of $N{a}_{2}S{O}_4$
$\alpha =\frac{i-1}{n-1}$
$\alpha =\frac{2.5-1}{3-1}$
$\alpha =\frac{1.5}{2}$
$\alpha =0.75=$ $75$%