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#nitrogen percentage

8 public questions tagged with this topic.

In the estimation of nitrogen by Dumas method, 0.4 g of a compound produces 84 mL of N₂ at STP, and in Carius method, 0.

Mass of N = (28/22400) × 84 = 0.105 g. Percentage = (0.105/0.4) × 100 = 26.25%. (Carius data is a distractor for chlorine.)

Ref: NCERT Class 11 Chemistry > Chapter 8: Organic Chemistry - Some Basic Principles and Techniques > Topic: Qualitative and Quantitative Analysis - Lassaigne Test and Estimation

In Kjeldahl’s method, 0.42 g of a compound produced ammonia that neutralized 15 mL of 0.2 M H₂SO₄. What is the percentag

Moles of H₂SO₄ = 0.015 × 0.2 = 0.003. Moles of NH₃ = 0.006 (2H per H₂SO₄). Mass of N = 0.006 × 14 = 0.084 g. Percentage = (0.084/0.42) × 100 = 20%.

Ref: NCERT Class 11 Chemistry > Chapter 8: Organic Chemistry - Some Basic Principles and Techniques > Topic: Isomerism - Stereoisomerism - Geometrical and Optical Isomerism

In Kjeldahl’s method, 0.36 g of a compound produced ammonia that neutralized 25 mL of 0.1 M H₂SO₄. What is the percentag

Moles of H₂SO₄ = 0.025 × 0.1 = 0.0025. Moles of NH₃ = 0.005 (2H per H₂SO₄). Mass of N = 0.005 × 14 = 0.07 g. Percentage = (0.07/0.36) × 100 ≈ 19.44%, which rounds to 19%.

Ref: NCERT Class 11 Chemistry > Chapter 8: Organic Chemistry - Some Basic Principles and Techniques > Topic: Electronic Effects - Inductive Mesomeric Hyperconjugation and Resonance