Passage: Consider the following reaction sequence in which J , K ,...

The reaction sequence proceeds as follows:

1. Friedel-Crafts Acylation: Reaction of $m$-xylene (1,3-dimethylbenzene) with chloroacetyl chloride ($\text{ClCH}_2\text{COCl}$) in the presence of anhydrous $\text{AlCl}_3$ occurs at the less sterically hindered activated position (position 4) to give 1-(2,4-dimethylphenyl)-2-chloroethan-1-one.

2. Finkelstein Reaction: Heating with $\text{NaI}$ replaces the chloride with iodide, yielding 1-(2,4-dimethylphenyl)-2-iodoethan-1-one.

3. Williamson Ether Synthesis: Reaction with sodium 3-nitrophenoxide gives compound $\textbf{J}$, which is 1-(2,4-dimethylphenyl)-2-(3-nitrophenoxy)ethan-1-one.

4. Reduction and Bromination: $\text{NaBH}_4$ reduces the ketone group in $\textbf{J}$ to a secondary alcohol without affecting the nitro group. Subsequent reaction with $\text{PBr}_3$ converts the alcohol to a bromide, yielding compound $\textbf{K}$, 1-(1-bromo-2-(3-nitrophenoxy)ethyl)-2,4-dimethylbenzene.
The molecular formula of $\textbf{K}$ is $\text{C}_{16}\text{H}_{16}\text{BrNO}_3$.
Molar mass of $\textbf{K} = (16 \times 12) + (16 \times 1) + 80 + 14 + (3 \times 16) = 192 + 16 + 80 + 14 + 48 = 350 \text{ g/mol}$, which matches the given data.

5. Amination: Reaction of $\textbf{K}$ with excess $\text{NH}_3$ results in nucleophilic substitution of the bromide to form a primary amine, compound $\textbf{L}$, 1-(2,4-dimethylphenyl)-2-(3-nitrophenoxy)ethanamine.
The molecular formula of $\textbf{L}$ is $\text{C}_{16}\text{H}_{18}\text{N}_2\text{O}_3$.
Molar mass of $\textbf{L} = (16 \times 12) + (18 \times 1) + (2 \times 14) + (3 \times 16) = 192 + 18 + 28 + 48 = 286 \text{ g/mol}$.

Moles of $\textbf{L}$ taken = $\dfrac{5.72 \text{ g}}{286 \text{ g/mol}} = 0.02 \text{ mol}$.

In Kjeldahl's method, nitrogen present in nitro ($-\text{NO}_2$) groups is not converted to ammonium sulfate and thus is not estimated. Only the nitrogen from the primary amine ($-\text{NH}_2$) group will evolve as ammonia ($\text{NH}_3$).
Therefore, 1 mole of $\textbf{L}$ yields 1 mole of $\text{NH}_3$.
Moles of $\text{NH}_3$ evolved = $0.02 \text{ mol}$.

The neutralization reaction with sulfuric acid is:
$2\text{NH}_3 + \text{H}_2\text{SO}_4 \rightarrow (\text{NH}_4)_2\text{SO}_4$

Moles of $\text{H}_2\text{SO}_4$ required = $\dfrac{0.02}{2} = 0.01 \text{ mol}$.

Given the molarity of $\text{H}_2\text{SO}_4$ is $1 \text{ M}$:
Volume of $\text{H}_2\text{SO}_4 = \dfrac{\text{Moles}}{\text{Molarity}} = \dfrac{0.01 \text{ mol}}{1 \text{ mol/L}} = 0.01 \text{ L} = 10 \text{ mL}$.

Answer: $10$