Exam

JIPMER

Year

2018

Subject

Chemistry

Chapter

Solutions

JIPMER 2018 · Chemistry · Solutions

The van't Hoff factor for 0.1 M Ba(NO_3)_2 solution is 2.74 . The...

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

Step-by-step 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 $=87$