Organic Chemistry: Some Basic Principles and Techniques — Important Questions
ICSE · Class 11 · Chemistry
63 important questions from Organic Chemistry: Some Basic Principles and Techniques for ICSE Class 11 Chemistry, with answers.
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Important Questions from Organic Chemistry: Some Basic Principles and Techniques
A 1.5 g sample of an organic compound gives 2.5090 g of Mg2P2O7. What is the percentage of phosphorus?
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46.71%
Percentage of phosphorus = (62 / 222) × (mass of Mg2P2O7 / mass of compound) × 100. Substituting gives (62 / 222) × (2.5090 / 1.5) × 100 = 46.71%.
A compound of mass 0.20 g is analysed by Liebig's method and gives 0.44 g of CO2 and 0.18 g of H2O. What are the percentages of carbon and hydrogen respectively?
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Carbon = 60.0%, Hydrogen = 10.0%
For carbon, %C = (12/44) × (mass of CO2 / mass of compound) × 100 = (12/44) × (0.44 / 0.20) × 100 = 60.0%. For hydrogen, %H = (2/18) × (mass of H2O / mass of compound) × 100 = (2/18) × (0.18 / 0.20) × 100 = 10.0%.
A 0.303 g organic compound is Kjeldahlised. The ammonia evolved is absorbed in 50 mL of 0.1 N H2SO4, and the excess acid requires 25 mL of 0.1 N NaOH. What is the percentage of nitrogen?
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11.55%
The acid neutralised by ammonia = 50 - 25 = 25 mL. Gram equivalents of NH3 = (0.1 × 25) / 1000. Mass of nitrogen = this equivalent × 14 = 0.035 g. Percentage nitrogen = (0.035 / 0.303) × 100 = 11.55%.
In Duma's method, 0.22 g of a compound gives 34 mL of moist nitrogen at 17 °C and 733.4 mm pressure. If aqueous tension at 17 °C is 13.4 mm, what is the percentage of nitrogen?
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17.23%
Dry nitrogen pressure = 733.4 - 13.4 = 720 mm. Convert 34 mL at 720 mm and 290 K to STP: V2 = (720 × 34 × 273) / (290 × 760) = 30.32 mL. Then %N = [28 × V(STP) × 100] / [22400 × mass] = [28 × 30.32 × 100] / [22400 × 0.22] = 17.23%.
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