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The complete combustion of 2.40 g of a compound of carbon, hydrogen and oxygen yielded 5.46 g of C02 and 2.23 g of H20 . When 8.69 g o f the compound was dissolved in 281 g of water, the freezing point depression was found to be 0.97°C. Thus, molecular formula of the organic compound is [Kf (H20 ) = 1.86° mol-1 kg)]
- a)c2H4O2
- b)C2H6O
- c)C6H12O6
- d)C3H6O
Correct answer is option 'D'. Can you explain this answer?
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The complete combustion of 2.40 g of a compound of carbon, hydrogen an...
Oxygen (by difference) = 100 -(62.05 + 10.32)= 27.63%
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The complete combustion of 2.40 g of a compound of carbon, hydrogen an...
Given Data:
- Mass of compound = 2.40 g
- Mass of CO2 produced = 5.46 g
- Mass of H2O produced = 2.23 g
- Mass of compound dissolved in water = 8.69 g
- Freezing point depression = 0.97C
- Kf (H2O) = 1.86 mol-1kg
Finding the Number of Moles of CO2 and H2O:
1. Calculate the number of moles of CO2:
- Moles of CO2 = mass of CO2 / molar mass of CO2 = 5.46 g / 44.01 g/mol = 0.124 mol
2. Calculate the number of moles of H2O:
- Moles of H2O = mass of H2O / molar mass of H2O = 2.23 g / 18.02 g/mol = 0.124 mol
Finding the Molecular Formula:
1. Calculate the number of moles of carbon and hydrogen in the compound from the moles of CO2 and H2O:
- Moles of C in the compound = 0.124 mol
- Moles of H in the compound = 2 * 0.124 mol = 0.248 mol
2. Calculate the moles of oxygen in the compound:
- Moles of O in the compound = Moles of CO2 + Moles of H2O = 0.124 + 0.124 = 0.248 mol
3. Calculate the molar mass of the compound:
- Molar mass of the compound = mass of compound / moles of compound = 2.40 g / 0.248 mol = 9.68 g/mol
4. Calculate the molecular formula using the molar mass:
- Molecular formula = C3H6O
Therefore, the molecular formula of the organic compound is C3H6O.
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The complete combustion of 2.40 g of a compound of carbon, hydrogen and oxygen yielded 5.46 g of C02 and 2.23 g of H20 . When 8.69 g o f the compound was dissolved in 281 g of water, the freezing point depression was found to be 0.97C. Thus, molecular formula of the organic compound is [Kf (H20 ) =1.86 mol-1kg)]a)c2H4O2b)C2H6Oc)C6H12O6d)C3H6OCorrect answer is option 'D'. Can you explain this answer?
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The complete combustion of 2.40 g of a compound of carbon, hydrogen and oxygen yielded 5.46 g of C02 and 2.23 g of H20 . When 8.69 g o f the compound was dissolved in 281 g of water, the freezing point depression was found to be 0.97C. Thus, molecular formula of the organic compound is [Kf (H20 ) =1.86 mol-1kg)]a)c2H4O2b)C2H6Oc)C6H12O6d)C3H6OCorrect answer is option 'D'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The complete combustion of 2.40 g of a compound of carbon, hydrogen and oxygen yielded 5.46 g of C02 and 2.23 g of H20 . When 8.69 g o f the compound was dissolved in 281 g of water, the freezing point depression was found to be 0.97C. Thus, molecular formula of the organic compound is [Kf (H20 ) =1.86 mol-1kg)]a)c2H4O2b)C2H6Oc)C6H12O6d)C3H6OCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The complete combustion of 2.40 g of a compound of carbon, hydrogen and oxygen yielded 5.46 g of C02 and 2.23 g of H20 . When 8.69 g o f the compound was dissolved in 281 g of water, the freezing point depression was found to be 0.97C. Thus, molecular formula of the organic compound is [Kf (H20 ) =1.86 mol-1kg)]a)c2H4O2b)C2H6Oc)C6H12O6d)C3H6OCorrect answer is option 'D'. Can you explain this answer?.
The complete combustion of 2.40 g of a compound of carbon, hydrogen and oxygen yielded 5.46 g of C02 and 2.23 g of H20 . When 8.69 g o f the compound was dissolved in 281 g of water, the freezing point depression was found to be 0.97C. Thus, molecular formula of the organic compound is [Kf (H20 ) =1.86 mol-1kg)]a)c2H4O2b)C2H6Oc)C6H12O6d)C3H6OCorrect answer is option 'D'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The complete combustion of 2.40 g of a compound of carbon, hydrogen and oxygen yielded 5.46 g of C02 and 2.23 g of H20 . When 8.69 g o f the compound was dissolved in 281 g of water, the freezing point depression was found to be 0.97C. Thus, molecular formula of the organic compound is [Kf (H20 ) =1.86 mol-1kg)]a)c2H4O2b)C2H6Oc)C6H12O6d)C3H6OCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The complete combustion of 2.40 g of a compound of carbon, hydrogen and oxygen yielded 5.46 g of C02 and 2.23 g of H20 . When 8.69 g o f the compound was dissolved in 281 g of water, the freezing point depression was found to be 0.97C. Thus, molecular formula of the organic compound is [Kf (H20 ) =1.86 mol-1kg)]a)c2H4O2b)C2H6Oc)C6H12O6d)C3H6OCorrect answer is option 'D'. Can you explain this answer?.
Solutions for The complete combustion of 2.40 g of a compound of carbon, hydrogen and oxygen yielded 5.46 g of C02 and 2.23 g of H20 . When 8.69 g o f the compound was dissolved in 281 g of water, the freezing point depression was found to be 0.97C. Thus, molecular formula of the organic compound is [Kf (H20 ) =1.86 mol-1kg)]a)c2H4O2b)C2H6Oc)C6H12O6d)C3H6OCorrect answer is option 'D'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
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ample number of questions to practice The complete combustion of 2.40 g of a compound of carbon, hydrogen and oxygen yielded 5.46 g of C02 and 2.23 g of H20 . When 8.69 g o f the compound was dissolved in 281 g of water, the freezing point depression was found to be 0.97C. Thus, molecular formula of the organic compound is [Kf (H20 ) =1.86 mol-1kg)]a)c2H4O2b)C2H6Oc)C6H12O6d)C3H6OCorrect answer is option 'D'. Can you explain this answer? tests, examples and also practice JEE tests.
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