Question

0.004 MNa$_2SO_4$ is isotonic with 0.01 M glucose. Degree of dissociation of Na$_2SO_4$ is

• A. 0.75
• B. 0.5
• C. 0.25
• D. 0.85

Answer 1

Answer: (A)

0.75

Answer 2

For isotonic solutions, they must have same concentrations of ions. Therefore,
$\, \, \, \, \, 0.004 i (Na_2SO_4)=0.01$
$\Rightarrow \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, i=\frac{0.01}{0.004}=2.5$
Also $\, \, Na_2SO_4 \rightleftharpoons 2 _{2 \alpha}^{Na^+}+ _{\alpha}^{SO_4^{2-}} \, _{1+2\alpha}^{i}$
$\Rightarrow \, \, \, \, \, \, \, \, \, \, \, i=1 + 2\alpha =2.5$
\hspace30mm \alpha =0.75 =75 \%