Dichloroacetic acid $\left({\mathrm{CHCl}}_2\mathrm{COOH}\right)$is oxidised to ${\mathrm{CO}}_2 , {\mathrm{H}}_2\mathrm{O}$ and ${\mathrm{Cl}}_2$ by 600 meq of an oxidising agent. Same amount of acid can neutralize how many moles of ammonia to form ammonium dichloroacetate :

Here, the reaction we have,

${\mathrm{CHCl}}_2\mathrm{COOH}⟶2\underset{+8}{{\mathrm{CO}}_2}+{\mathrm{H}}_2\mathrm{O}+\underset{0}{{\mathrm{Cl}}_2}$

or $\begin{array}{c}{\mathrm{C}}_2{\mathrm{H}}_2{\mathrm{O}}_2{\mathrm{Cl}}_2\\ -4+2-4-2\end{array}$

So, the n factor = 6

$\mathrm{m} \mathrm{mol}\text{ of }{\mathrm{CHCl}}_2\mathrm{COOH}=\frac{600}{6}=100$

As an acid $\mathrm{n}.\mathrm{f}.=1$

So, m mol of salt = 100

Thus, the moles of ammonia required is 0.1.