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then the solubility of $AB _{2}$ in pure water is......... $\times 10^{-4}$ $mol$ $L ^{-1}$.
[Assuming that neither kind of ion reacts with water]
$(i)\,\mathop {\begin{array}{*{20}{c}}
{\,\,\,\,\,C{H_3}} \\
| \\
{C{H_3} - C - C{H_3}} \\
| \\
{\,\,\,\,\,C{H_3}}
\end{array}}\limits_{(Neo - pen\tan e)\,(i)} $
$(ii\mathop {)\,\begin{array}{*{20}{c}}
{C{H_3} - CH - C{H_2} - C{H_3}} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{\,\,\,\,\,C{H_{3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}}}
\end{array}}\limits_{(Iso - pen\tan e)\,\,(ii)} $
$(iii)\,\mathop {C{H_3} - C{H_2} - C{H_2} - C{H_2} - C{H_3}}\limits_{(n - pen\tan e)} $
$ \mathrm{N}_2=3.0 \times 10^{-3} \mathrm{M}, \mathrm{O}_2=4.2 \times 10^{-3} \mathrm{M} \text { and } \mathrm{NO}=2.8 \times 10^{-3} \mathrm{M} . $
$ 2 \mathrm{NO}_{(\mathrm{g})} \rightleftharpoons \mathrm{N}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})}$
If $0.1 \mathrm{~mol} \mathrm{~L} \mathrm{~L}^{-1}$ of $\mathrm{NO}_{(\mathrm{g})}$ is taken in a closed vessel, what will be degree of dissociation ( $\alpha$ ) of $\mathrm{NO}_{(\mathrm{g})}$ at equilibrium?