Consider the following reaction sequence in which J, K, L and M are the major products.

Given:
Atomic mass (in amu): H : 1, C : 12, N : 14, O : 16, S : 32, Br : 80, Ba : 137

The volume of 1 M aqueous H₂SO₄ required to completely neutralize the ammonia evolved from 5.72 g of L in Kjeldahl's method of nitrogen estimation is _______ mL.

The correct answer is 10.

Let us analyze the reaction sequence step-by-step to identify the chemical structures of compound J, K, and L.

Step 1: Formation of Compound J
Starting material is m-xylene (1,3-dimethylbenzene).
1. Friedel-Crafts acylation of m-xylene with chloroacetyl chloride (Cl-COCH₂Cl) in the presence of anhydrous AlCl₃ introduces a -COCH₂Cl group at position 4 (ortho to one methyl and para to the other methyl group, which is the most activated site).
2. Reaction with NaI and heating converts the alkyl chloride to alkyl iodide via Finkelstein reaction, giving 4-(2-iodoacetyl)-1,3-dimethylbenzene.
3. Williamson ether synthesis with sodium 3-nitrophenolate (NaO-C₆H₄-NO₂) replaces the iodide with 3-nitrophenoxy group to give compound J:
Compound J: 2-(3-nitrophenoxy)-1-(2,4-dimethylphenyl)ethan-1-one.

Step 2: Formation of Compound K
1. NaBH₄ reduces the ketone carbonyl group (-C=O) of J into a secondary alcohol (-CH(OH)-).
2. Reaction with PBr₃ replaces the hydroxyl group (-OH) with a bromine atom (-Br), forming compound K.
Chemical formula of K: C₁₆H₁₆BrNO₃.
Molar mass of K = 16(12) + 16(1) + 80 + 14 + 3(16) = 192 + 16 + 80 + 14 + 48 = 350 g/mol (which matches the given molar mass of 350 g/mol).

Step 3: Formation of Compound L
Reaction of compound K with excess ammonia (NH₃) results in nucleophilic substitution (S_N2) of the bromine atom by an amino group (-NH₂).
Chemical formula of compound L: C₁₆H₁₈N₂O₃.
Let us calculate the molar mass of L:
Molar mass of L = 16(12) + 18(1) + 2(14) + 3(16)
= 192 + 18 + 28 + 48 = 286 g/mol.

Step 4: Kjeldahl's Method and Neutralization Calculation
Mass of L taken = 5.72 g.
Number of moles of L = $\frac{5.72}{286}=0.02\text{ mol}$

Each molecule of compound L contains 2 nitrogen atoms (one from the nitro group -NO₂ and one from the amine group -NH₂). In standard Kjeldahl's method, nitrogen present in nitro (-NO₂) groups is not converted into ammonia unless prior reduction is mentioned; however, in standard stoichiometric Kjeldahl problems where compound L undergoes estimation of amine nitrogen or total nitrogen converted to ammonia:
Number of moles of NH₃ evolved = 0.02 mol.

Reactions involved in neutralization with aqueous H₂SO₄:
2 NH₃ + H₂SO₄ → (NH₄)₂SO₄

According to stoichiometry:
Moles of H₂SO₄ required = $\frac{\text{Moles of }{\text{NH}}_3}{2}=\frac{0.02}{2}=0.01\text{ mol}$

Given molarity of H₂SO₄ = 1 M (1 mol/L).
Volume of H₂SO₄ in liters = $\frac{\text{Moles}}{\text{Molarity}}=\frac{0.01}{1}=0.01\text{ L}$

Converting volume to mL:
Volume in mL = 0.01 × 1000 = 10 mL.

Thus, the volume of 1 M aqueous H₂SO₄ required is 10 mL.