Solutions and Colligative Properties - Result Question 20-1
20. $0.004$ $ \mathrm{M} $ $\mathrm{Na}_2 \mathrm{SO}_4$ is isotonic with $0.01 $ M glucose. Degree of dissociation of $\mathrm{Na}_2 \mathrm{SO}_4$ is
(2004, S, 1M)
(a) $75 %$
(b) $50 %$
(c) $25 %$
(d) $85 %$
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Answer:
Correct Answer: 20. (a)
Solution:
For isotonic solutions, they must have same concentrations of ions, Therefore,
$ 0.004 $ $i\left(\mathrm{Na}_2 \mathrm{SO}_4\right)=0.01 $
$ \Rightarrow \quad i=\dfrac{0.01}{0.004}=2.5 $
Also $ \underset{1-\alpha}{\mathrm{Na}_2 \mathrm{SO}_4} \rightleftharpoons \underset{2 \alpha}{2 Na^+}$ $+\underset{\alpha}{\mathrm{SO}_4^{2-}} \quad\underset{1 + 2 \alpha}{i} $
$ \Rightarrow \quad i=1+2 \alpha=2.5 $
$ \alpha=0.75=75 % $
BETA
