Question

If a 0.004 M solution of $Na_{2}SO_{4}$ is isotonic with a 0.010 M solution of glucose at same temperature. Then what is the apparent degree of dissociation of $Na_{2}SO_{4}$ is

• A. 25%
• B. 50%
• C. 75%
• D. 85%

Answer 1

Answer: (C)

Isotonic solutions have equal osmotic pressures.
$\underset{\left(\right. 0.004 - x \left.\right)}{N a_{2} S O_{4}}\rightleftharpoons\underset{2 x}{2 N a^{+}}+\underset{x}{S O_{4}^{2 -}}$
$\pi _{1}=\pi _{2}i_{1}\times C_{1}\times RT=i_{2}\times C_{2}\times RTi_{1}\times C_{1}=i_{2}\times C_{2}i_{1}\times 0.004=1\times 0.01i_{1}=2.5i_{1}=1+2\alpha \alpha =\frac{2 . 5 - 1}{2}=0.75\%\alpha =0.75\times 100=75\%$
or
Since both the solution are isotonic $0.004+2x=0.01$
$x=0.003$
Percent dissociation of $Na_{2}SO_{4}=\frac{3 \times 10^{- 3}}{0 .004}\times 100=75\%$ .