1. In Carius estimation, 0.2 g of an organic compound gave 0.404 g of AgCl. The percentage of chlorine in the compound is:
- 16.4 %
- 100 %
- 25 %
- 50 %
Show answer and explanation
Answer: (D) 50 %
%chlorine = (35.5/143.5) × (0.404/0.2) × 100 = 50%.
90 questions on this topic of Purification and Characterisation of Organic Compounds (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.
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Answer: (D) 50 %
%chlorine = (35.5/143.5) × (0.404/0.2) × 100 = 50%.
Answer: (B) 40 %
%bromine = (80/188) × (0.235/0.25) × 100 = 40%.
Answer: (D) 30 %
%C = (12/44) × (mass of CO₂/mass of compound) × 100 = (12/44) × (0.55/0.5) × 100 = 30%.
Answer: (C) 55 %
%chlorine = (35.5/143.5) × (0.333/0.15) × 100 = 55%.
Answer: (A) 4.67 %
%N = 1.4 × N × V/m = 1.4 × 0.1 × 25/0.75 = 4.67%.
Answer: (A) 35 %
%bromine = (80/188) × (0.123/0.15) × 100 = 35%.
Answer: (D) 85 %
%C = (12/44) × (mass of CO₂/mass of compound) × 100 = (12/44) × (0.623/0.2) × 100 = 85%.
Answer: (B) 50 %
%C = (12/44) × (mass of CO₂/mass of compound) × 100 = (12/44) × (0.458/0.25) × 100 = 50%.
Answer: (A) 56 %
%N = 1.4 × N × V/m = 1.4 × 1 × 10/0.25 = 56%.
Answer: (C) 70 %
%C = (12/44) × (mass of CO₂/mass of compound) × 100 = (12/44) × (0.77/0.3) × 100 = 70%.
Answer: (C) 30 %
%bromine = (80/188) × (0.106/0.15) × 100 = 30%.
Answer: (D) 75 %
%C = (12/44) × (mass of CO₂/mass of compound) × 100 = (12/44) × (0.55/0.2) × 100 = 75%.