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Chemistry Question 74 – JEE-MAIN 2025

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X g of nitrobenzene on nitration gave 4.2 g of m-dinitrobenzene. X = _______ g. (nearest integer) [Given : molar mass (in g mol−1) C : 12, H : 1, O : 16, N : 14]

Identify the chemical reaction occurring (nitration of nitrobenzene) and the product formed (m-dinitrobenzene).

Step 1: Identify Reactant, Product, and their Molar Masses✦ Active

The reaction is the nitration of nitrobenzene (C6H5NO2) to form m-dinitrobenzene (C6H4(NO2)2). The nitro group is meta-directing, so the product is indeed m-dinitrobenzene. The balanced reaction is:

C6H5NO2→NitrationC6H4(NO2)2

The stoichiometric ratio between nitrobenzene and m-dinitrobenzene is 1:1. Now, calculate their molar masses:

Molar mass of nitrobenzene (C6H5NO2)=(6×12)+(5×1)+(1×14)+(2×16)=72+5+14+32=123 g/mol
Molar mass of m-dinitrobenzene (C6H4(NO2)2)=(6×12)+(4×1)+(2×14)+(4×16)=72+4+28+64=168 g/mol
Step 2: Calculate Moles of Product○ Expand

Given that 4.2 g of m-dinitrobenzene is formed, we can calculate the moles of product:

Moles of m-dinitrobenzene=MassMolar Mass=4.2 g168 g/mol=0.025 mol
Step 3: Calculate Mass of Reactant (X)○ Expand

Since the stoichiometric ratio between nitrobenzene and m-dinitrobenzene is 1:1, the moles of nitrobenzene reacted must also be 0.025 mol. Now, calculate the mass of nitrobenzene (X):

Mass of nitrobenzene (X)=Moles×Molar Mass=0.025 mol×123 g/mol=3.075 g

Rounding to the nearest integer, X = 3 g.

💡 Teacher's Secret Hint

Ensure to round the final answer to the nearest integer as specified in the question.

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