Question

If a 0.004 M solution of $N{a}_{2}S{O}_4$ is isotonic with a 0.010 M solution of glucose at same temperature. Then what is the apparent degree of dissociation of $N{a}_{2}S{O}_4$ is

• A. 25%
• B. 50%
• C. 75%
• D. 85%

Answer 1

Answer: (C)

Isotonic solutions have equal osmotic pressures.
$\underset{(0.004-x)}{N{a}_{2}S{O}_4}\rightleftharpoons \underset{2x}{2N{a}^{+}}+\underset{x}{S{O}_4^{2-}}$
${\pi }_1={\pi }_2{i}_1\times C_1\times RT=i_2\times C_2\times RT{i}_1\times C_1=i_2\times C_2{i}_1\times 0.004=1\times 0.01i_1=2.5i_1=1+2\alpha \alpha =\frac{2.5-1}{2}=0.75\%\alpha =0.75\times 100=75\%$
or
Since both the solution are isotonic $0.004+2x=0.01$
$x=0.003$
Percent dissociation of $N{a}_{2}S{O}_4=\frac{3\times 10^{-3}}{0.004}\times 100=75\%$ .