Q62 / 180 · JIPMER 2018 · Chemistry
The van't Hoff factor for $0.1 \mathrm{M} \mathrm{Ba}\left(\mathrm{NO}_3\right)_2$ solution is 2.74 . The degree of dissociation is
- A$91.3 \%$
- Correct answerB$87 \%$
- C$100 \%$
- D$74 \%$
Solution
$\mathrm{Ba}\left(\mathrm{NO}_3\right)_2 \rightleftharpoons \mathrm{Ba}^{2+}+2 \mathrm{NO}_3^{-}$
Total number of moles $=1-\alpha+\alpha+2 \alpha$
$=1+2 \alpha$
Then, van't Hoff factor, $\mathrm{i}=1+2 \alpha$
or, $\quad \alpha=\frac{i-1}{2}$
or, $\quad \alpha=\frac{2.74-1}{2}=0.87 \%$
$\%$ dissociation $=87$
Total number of moles $=1-\alpha+\alpha+2 \alpha$
$=1+2 \alpha$
Then, van't Hoff factor, $\mathrm{i}=1+2 \alpha$
or, $\quad \alpha=\frac{i-1}{2}$
or, $\quad \alpha=\frac{2.74-1}{2}=0.87 \%$
$\%$ dissociation $=87$Hence, the correct answer is option: 'B'.