JEE Main 2025 • Chemistry • Numerical Answer
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
Exam: JEE Main 2025 (Online)
Shift: 3rd April Evening Shift
Step by Step Solution
Step 1:
Write the molecular formula of nitrobenzene.
Nitrobenzene = C₆H₅NO₂
Step 2:
Calculate molar mass of nitrobenzene.
= (6 × 12) + (5 × 1) + (1 × 14) + (2 × 16)
= 72 + 5 + 14 + 32
= 123 g mol⁻¹
Step 3:
On nitration, nitrobenzene forms m-dinitrobenzene.
Molecular formula of m-dinitrobenzene = C₆H₄N₂O₄
Step 4:
Calculate molar mass of m-dinitrobenzene.
= (6 × 12) + (4 × 1) + (2 × 14) + (4 × 16)
= 72 + 4 + 28 + 64
= 168 g mol⁻¹
Step 5:
From stoichiometry,
1 mole of nitrobenzene → 1 mole of m-dinitrobenzene
Step 6:
Moles of m-dinitrobenzene formed = 4.2 / 168
= 0.025 moles
Step 7:
Same moles of nitrobenzene were required.
Mass of nitrobenzene = 0.025 × 123
= 3.075 g
Step 8:
Nearest integer value of X = 3 g
✅ Final Answer: 3 g