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
In sulphur estimation by Carius method, the amount of formed from 3.79 g of is ____ g.
Given:
Atomic mass (in amu): H : 1, C : 12, N : 14, O : 16, S : 32, Br : 80, Ba : 137
In sulphur estimation by Carius method, the amount of formed from 3.79 g of is ____ g.
Answer: 2.33
Solution
From the reaction sequence, compound M has molecular formula C₂₂H₂₁NO₃S with molar mass 379 g/mol.
Moles of M = 3.79/379 = 0.01 mol
Each molecule of M contains 1 sulfur atom.
In Carius method, sulfur is oxidized to SO₄²⁻ and precipitated as BaSO₄.
Moles of BaSO₄ formed = 0.01 mol
Molar mass of BaSO₄ = 137 + 32 + 4(16) = 233 g/mol
Mass of BaSO₄ = 0.01 × 233 = 2.33 g
Moles of M = 3.79/379 = 0.01 mol
Each molecule of M contains 1 sulfur atom.
In Carius method, sulfur is oxidized to SO₄²⁻ and precipitated as BaSO₄.
Moles of BaSO₄ formed = 0.01 mol
Molar mass of BaSO₄ = 137 + 32 + 4(16) = 233 g/mol
Mass of BaSO₄ = 0.01 × 233 = 2.33 g
JEE Advanced · 2026 · 17 May 2026 · Paper 2 · Numerical answer