Chemistry · Nitrogen Containing Organic Compounds

JEE Main 2024 — 6 April, Shift 1 — Question 79

$9.3\ \mathrm{g}$ of pure aniline upon diazotisation followed by coupling with phenol gives an orange dye. The mass of orange dye produced (assuming $100\%$ yield) is $\_\_\_$ g (nearest integer).

Answer: 20

Numerical answer — enter this value.

Step-by-step solution

Molecular mass of aniline $(\mathrm{C_6H_5NH_2})$ :

M = 6(12) + 5(1) + 14 + 2(1) = 93\ \mathrm{g\,mol^{-1}}

Moles of aniline $= n = \frac{9.3}{93} = 0.1\ \mathrm{mol}$

Diazotisation and coupling with phenol proceed 1:1 giving one mole of azo dye per mole of aniline.

Molecular mass of the azo dye (p-hydroxyazobenzene) $\mathrm{C_{12}H_{10}N_2O}$ is

M = 12(12) + 10(1) + 2(14) + 16 = 144 + 10 + 28 + 16 = 198\ \mathrm{g\,mol^{-1}}

Mass of dye formed $= m = n \times M = 0.1 \times 198 = 19.8\ \mathrm{g}$ Nearest integer = $20$ .

Answer key and solution verified before publishing.

Practise Nitrogen Containing Organic Compounds

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Main 2024
Paper: 6 April, Shift 1
Subject: Chemistry
Chapter: Nitrogen Containing Organic Compounds
Topic: Diazonium salts