- A${N_2} \to 2N{H_3}$
- B$N_2O_4 \to 2NO_2$
- C$NO_3^- \to N_2O_5$
- ✓$NO_2^- \to NO_3^-$
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Given : $K_b(H_2O) = 0.51\, Kkg \,mol^{-1}$
$N_A = 6 × 10^{23}$
$[Li = 7, \,\,Na = 23$, $K = 39,\,\, Rb = 85.5$ , $Cs = 133, \,\,Cl = 35.5]$

$\mathop {\begin{array}{*{20}{c}}
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,O}\\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,||}\\
{CH \equiv C - C - OH}
\end{array}}\limits_{\rm{I}} $ $\mathop {\begin{array}{*{20}{c}}
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,O}\\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,||}\\
{C{H_2} = CH - C - OH}
\end{array}}\limits_{{\rm{II}}} $ $\mathop {\begin{array}{*{20}{c}}
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,O}\\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,||}\\
{C{H_3} - C{H_2} - C - OH}
\end{array}}\limits_{{\rm{III}}} $$\mathop {\begin{array}{*{20}{c}}
\,\,\,{C{H_3}\,\,\,\,\,\,O}\\
\,\,\,{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,||}\\
{C{H_3} - CH - C - OH}
\end{array}}\limits_{{\rm{IV}}} $