Heat liberated in the neutralization of $500\mathrm{ml}$ of $1\mathrm{ }\mathrm{N}\mathrm{HCl}$ and $500\mathrm{ml}$ of $1\mathrm{ }\mathrm{N}{\mathrm{NH}}_4\mathrm{OH}$ is $1.36$ K.Cals. The heat of ionization of ${\mathrm{NH}}_4\mathrm{OH}$ is

$\mathrm{NaOH}+\mathrm{HCl}\to \mathrm{NaCl}+{\mathrm{H}}_2\mathrm{O} \Delta \mathrm{H}=-13.7\mathrm{kcal}/\mathrm{eq}$

As we know the heat of neutralization of a strong acid by a strong base is the heat of formation of water.

Therefore,

${\mathrm{H}}^{+}+{\mathrm{OH}}^{-}\to {\mathrm{H}}_2\mathrm{O} \Delta \mathrm{H}=-13.7\mathrm{kcal}$.

${\mathrm{H}}_2\mathrm{O}\to {\mathrm{H}}^{+}+{\mathrm{OH}}^{-} \Delta \mathrm{H}=13.7\mathrm{kcal}\dots ..\left(1\right)$.

Given that heat liberated in the neutralization of $500\mathrm{ }\mathrm{m}\mathrm{L}$ of $1\mathrm{ }\mathrm{N}\mathrm{HCl}$ and $500\mathrm{ }\mathrm{m}\mathrm{L}$ of $1\mathrm{ }\mathrm{N}{\mathrm{NH}}_4\mathrm{OH}$ is $-1.36\mathrm{kcal}$

As we know,

$gmeq.=\mathrm{N}\times {\mathrm{V}}_{\left(\mathrm{inL}\right)}$

$\therefore$ gm eq:HCl $=$ gmeq ${}_{{\mathrm{NH}}_4\mathrm{OH}}=1\times \frac{500}{1000}=0.5$

Now,

The heat liberated in the neutralization of $0.5$ gram equivalent of $\mathrm{HCl}$ and

${\mathrm{NH}}_4\mathrm{OH}=-1.36\mathrm{kcal}$

The heat liberated in the neutralization of 1 gram equivalent of $\mathrm{HCl}$ and

${\mathrm{NH}}_4\mathrm{OH}=\frac{-1.36}{0.5}=-2.72\mathrm{kcal}$

Therefore,

${\mathrm{H}}^{+}+{\mathrm{NH}}_4\mathrm{OH}\to {\mathrm{NH}}_4^{+}+{\mathrm{H}}_2\mathrm{O} \Delta \mathrm{H}=-2.72\mathrm{kcal}.$

Adding eq" (1) &  (2), we have

$$\begin{gathered} {\mathrm{H}}_2\mathrm{O}+{\mathrm{NH}}_4\mathrm{OH}+{\mathrm{H}}^{+}\to {\mathrm{H}}^{+}+{\mathrm{OH}}^{-}+{\mathrm{NH}}_4{}^{-}+{\mathrm{H}}_2\mathrm{O} \Delta \mathrm{H}=13.7+(-2.72) \\ {\mathrm{NH}}_4\mathrm{OH}\to {\mathrm{NH}}_4{}^{+}+{\mathrm{OH}}^{-} \Delta \mathrm{H}=10.98\mathrm{kcal} \end{gathered}$$

Hence the heat of ionization of ${\mathrm{NH}}_4\mathrm{OH}$ is $10.98\mathrm{kcal}$.

Hence option A  is the correct answer.