Equilibrium JEE 2025: Nylon 6,6 Dumas Nitrogen Answer

What is the answer to the Nylon 6,6 Dumas question in Equilibrium JEE 2025?

The answer to the Nylon 6,6 numerical listed under “equilibrium jee 2025” is 280.00 g of nitrogen gas. It is from JEE Advanced 2025, Paper 1, Chemistry, in numerical-answer format. The question bank’s medium difficulty tag and 90-second expected solving time are bank metadata, not official exam classifications.

The question asks: Of the monomers used to make Nylon 6,6, monomer X gives a positive carbylamine test. Determine the mass, in grams, of nitrogen gas evolved when 10 moles of X undergo Dumas nitrogen estimation. Take the atomic mass of nitrogen as 14 amu.

The bank classifies this under Equilibrium, but the solution uses polymer chemistry, the carbylamine test and stoichiometry. No equilibrium constant or equilibrium table is needed. The decisive step is distinguishing nitrogen atoms from nitrogen-gas molecules.

Which Nylon 6,6 monomer gives a positive carbylamine test?

Hexamethylenediamine is the required monomer, because it contains primary amino groups. Nylon 6,6 forms by condensation polymerisation of adipic acid and hexamethylenediamine. The official solution uses the carbylamine clue to choose between them before counting nitrogen.

Adipic acid has two terminal carboxylic acid groups, as its structure shows: $H O O C (C H_{2})_{4} C O O H$

The carbylamine test identifies primary amines, so adipic acid does not satisfy the clue. The required monomer is:

X = H_{2} N (C H_{2})_{6} N H_{2}

Hexamethylenediamine has two terminal primary amino groups. Each nitrogen is attached to one carbon atom and two hydrogen atoms, so each amino group is primary.

Two amino groups do not make this compound a secondary amine. “Primary” describes the bonding around each nitrogen; “diamine” counts the amino groups in the molecule. Therefore, each molecule of X contains exactly two nitrogen atoms.

How do 10 moles of the monomer give 280.00 g of nitrogen?

One mole of hexamethylenediamine gives one mole of nitrogen gas, not two. Under the Dumas-method accounting used here, all nitrogen in the organic compound is recovered as nitrogen gas. Each monomer molecule supplies two nitrogen atoms, and each nitrogen-gas molecule contains two nitrogen atoms.

Start by counting nitrogen atoms:

1 m o l X contains 2 m o l N a t o m s

For the given sample:

n (N a t o m s) = 10 m o l X \times \frac{2 m o l N a t o m s}{1 m o l X} = 20 m o l N a t o m s

Divide by two to convert nitrogen atoms into diatomic gas molecules:

n (N_{2}) = 20 m o l N a t o m s \times \frac{1 m o l N_{2}}{2 m o l N a t o m s} = 10 m o l N_{2}

The resulting molar relationship is:

1 m o l h e x a m e t h y l e n e d i a m i n e ⟶ 1 m o l N_{2}

Use the supplied atomic mass to calculate the gas molar mass. Then multiply by the moles of gas:

M (N_{2}) = 2 \times 14 = 28 g m o l^{- 1}

m (N_{2}) = 10 m o l \times 28 g m o l^{- 1} = 280 g

m (N_{2}) = 280.00 g

Gas pressure, temperature and molar volume are unnecessary because the requested quantity is mass, not volume. Keep the complete audit trail:

10 m o l X ⟶ 20 m o l N a t o m s ⟶ 10 m o l N_{2} ⟶ 280 g

Why would 560 g be a counting error?

560 g would result from treating moles of nitrogen atoms as moles of nitrogen gas. This is a numerical-answer question, and the supplied record contains no multiple-choice options. The result below is a hypothetical incorrect answer, not an official distractor.

The incorrect chain starts with two nitrogen atoms per monomer molecule, then wrongly treats that as two gas molecules:

10 m o l X \overset{incorrect}{\to} 20 m o l N_{2}

m_{wrong} = 20 m o l \times 28 g m o l^{- 1} = 560 g

The count of 20 moles refers to nitrogen atoms. Converting it to nitrogen-gas molecules requires division by two.

Atom conservation exposes the mistake:

20 m o l N_{2} contains 40 m o l N a t o m s

The sample contains only 20 moles of nitrogen atoms, so that gas amount is impossible. Label every amount as moles of monomer, moles of nitrogen atoms or moles of nitrogen gas, never as an unlabelled mole count.

Which two practice questions actually require equilibrium calculations?

The ammonia-formation and dinitrogen-tetroxide questions below require equilibrium calculations. They are author-created chapter practice questions, not additional JEE 2025 PYQs. Both require careful species counting, but unlike Dumas mass accounting, they explicitly involve reversible reactions and equilibrium concentrations.

How do I calculate the concentration equilibrium constant for ammonia formation?

Use the equilibrium concentrations, with balanced coefficients as powers. Under the standard JEE concentration convention, the answer is:

K_{c} = 0.50 L^{2} m o l^{- 2}

Question 1: For the reaction below, calculate the concentration equilibrium constant using the supplied equilibrium concentrations.

N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)

[N_{2}] = 1.0 m o l L^{- 1}, [H_{2}] = 2.0 m o l L^{- 1}, [N H_{3}] = 2.0 m o l L^{- 1}

Worked answer: All three concentrations are already equilibrium values. Substitute them directly:

K_{c} = \frac{[N H_{3}]^{2}}{[N_{2}] [H_{2}]^{3}} = \frac{2^{2}}{1 \times 2^{3}} = 0.50 L^{2} m o l^{- 2}

How do I calculate equilibrium concentrations after dinitrogen tetroxide dissociates?

Subtract the dissociated amount from the reactant and form twice that amount of nitrogen dioxide. Divide each amount by the vessel volume to obtain:

[N_{2} O_{4}] = 0.80 m o l L^{- 1}, [N O_{2}] = 0.40 m o l L^{- 1}

Question 2: A 1.0 L vessel initially contains 1.0 mol of dinitrogen tetroxide and no nitrogen dioxide. At equilibrium, 0.20 mol of dinitrogen tetroxide has dissociated through the following reaction. Calculate both equilibrium concentrations and the concentration equilibrium constant.

N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)

Worked answer:

n (N_{2} O_{4}) = 1.0 - 0.20 = 0.80 m o l

n (N O_{2}) = 2 \times 0.20 = 0.40 m o l

[N_{2} O_{4}] = \frac{0.80}{1.0} = 0.80 m o l L^{- 1}, [N O_{2}] = \frac{0.40}{1.0} = 0.40 m o l L^{- 1}

K_{c} = \frac{[N O_{2}]^{2}}{[N_{2} O_{4}]} = \frac{{0.40}^{2}}{0.80} = 0.20 m o l L^{- 1}

Before using any factor of two, name its source. In Question 2, multiplication comes from the balanced reaction; in Dumas accounting, division converts nitrogen atoms into nitrogen-gas molecules.

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Frequently asked questions

What is the answer to the JEE Advanced 2025 Nylon 6,6 Dumas question?

The answer is 280.00 g of nitrogen gas. Ten moles of hexamethylenediamine contain 20 moles of nitrogen atoms, which give 10 moles of N2. Multiplying by the molar mass of N2, 28 g/mol, gives 280.00 g.

Which Nylon 6,6 monomer gives a positive carbylamine test?

Hexamethylenediamine, H2N(CH2)6NH2, gives a positive carbylamine test because it contains primary amino groups. The other monomer, adipic acid, contains carboxylic acid groups and does not satisfy the clue. Having two amino groups makes hexamethylenediamine a diamine, not a secondary amine.

Why is 560 g wrong in the Nylon 6,6 Dumas question?

The value 560 g comes from incorrectly treating 20 moles of nitrogen atoms as 20 moles of N2. Since each N2 molecule contains two nitrogen atoms, 20 moles of nitrogen atoms form only 10 moles of N2. The correct mass is therefore 10 × 28 = 280 g.

Does the Nylon 6,6 Dumas question need an equilibrium calculation?

No equilibrium constant or equilibrium table is needed, despite the question bank listing it under Equilibrium. The solution uses Nylon 6,6 monomer identification, the carbylamine test and nitrogen stoichiometry. The article's ammonia and dinitrogen tetroxide exercises do require equilibrium calculations, but they are author-created practice questions, not additional JEE 2025 PYQs.

Equilibrium JEE 2025: Nylon 6,6 Dumas Nitrogen Answer

What is the answer to the Nylon 6,6 Dumas question in Equilibrium JEE 2025?

Which Nylon 6,6 monomer gives a positive carbylamine test?

How do 10 moles of the monomer give 280.00 g of nitrogen?

Why would 560 g be a counting error?

Which two practice questions actually require equilibrium calculations?

How do I calculate the concentration equilibrium constant for ammonia formation?

How do I calculate equilibrium concentrations after dinitrogen tetroxide dissociates?

Frequently asked questions

What is the answer to the JEE Advanced 2025 Nylon 6,6 Dumas question?

Which Nylon 6,6 monomer gives a positive carbylamine test?

Why is 560 g wrong in the Nylon 6,6 Dumas question?

Does the Nylon 6,6 Dumas question need an equilibrium calculation?

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