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A 100 mL mixture of CO and CO2 is passed through a tube containing red hot charcoal. The volume now becomes 160 mL. The volumes are measured under the same conditions of temperature and pressure. Amongst the following, select the correct statement (s).
- a)The mixture contains 40 mL of CO2
- b)Mole percent of CO2 in the mixture is 60.
- c)Mole fraction of CO in the mixture is 0.40
- d)The mixture contains 40 mL of CO
Correct answer is option 'B,C,D'. Can you explain this answer?
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Most Upvoted Answer
A 100 mL mixture of CO and CO2 is passed through a tube containing re...
Given Information:
- Initial volume of the mixture = 100 mL
- Final volume of the mixture after passing through the tube = 160 mL
- The volumes are measured under the same conditions of temperature and pressure.
Explanation:
To solve this problem, we need to understand the reaction that occurs when the mixture of CO and CO2 is passed through red hot charcoal.
When carbon monoxide (CO) is passed over red hot charcoal, it reacts with the charcoal to form carbon dioxide (CO2):
CO + C → CO2
The reaction consumes one molecule of CO and produces one molecule of CO2. Therefore, the total number of molecules remains the same before and after the reaction. This means that the moles of CO and CO2 also remain the same.
Now, let's analyze the given options:
a) The mixture contains 40 mL of CO2:
Since the total volume of the mixture increased from 100 mL to 160 mL, it implies that the reaction consumed some of the CO gas and produced an equal volume of CO2 gas. Therefore, the mixture does contain 40 mL of CO2.
b) Mole percent of CO2 in the mixture is 60:
To calculate the mole percent of CO2, we need to find the moles of CO and CO2 in the mixture. Since the moles of CO and CO2 remain the same before and after the reaction, the mole percent of CO2 in the mixture is given by:
Mole percent of CO2 = (moles of CO2 / total moles) * 100
Since the total moles remain the same, the mole percent of CO2 will also remain the same. Therefore, the mole percent of CO2 in the mixture is indeed 60.
c) Mole fraction of CO in the mixture is 0.40:
To calculate the mole fraction of CO, we need to find the moles of CO and CO2 in the mixture. Since the moles of CO and CO2 remain the same before and after the reaction, the mole fraction of CO is given by:
Mole fraction of CO = moles of CO / total moles
Since the total moles remain the same, the mole fraction of CO will remain the same. Therefore, the mole fraction of CO in the mixture is indeed 0.40.
d) The mixture contains 40 mL of CO:
Since the total volume of the mixture increased from 100 mL to 160 mL, it implies that the reaction consumed some of the CO gas. Therefore, the mixture does not contain 40 mL of CO.
Conclusion:
Based on the analysis, the correct statements are:
- The mixture contains 40 mL of CO2
- The mole percent of CO2 in the mixture is 60
- The mole fraction of CO in the mixture is 0.40
- Initial volume of the mixture = 100 mL
- Final volume of the mixture after passing through the tube = 160 mL
- The volumes are measured under the same conditions of temperature and pressure.
Explanation:
To solve this problem, we need to understand the reaction that occurs when the mixture of CO and CO2 is passed through red hot charcoal.
When carbon monoxide (CO) is passed over red hot charcoal, it reacts with the charcoal to form carbon dioxide (CO2):
CO + C → CO2
The reaction consumes one molecule of CO and produces one molecule of CO2. Therefore, the total number of molecules remains the same before and after the reaction. This means that the moles of CO and CO2 also remain the same.
Now, let's analyze the given options:
a) The mixture contains 40 mL of CO2:
Since the total volume of the mixture increased from 100 mL to 160 mL, it implies that the reaction consumed some of the CO gas and produced an equal volume of CO2 gas. Therefore, the mixture does contain 40 mL of CO2.
b) Mole percent of CO2 in the mixture is 60:
To calculate the mole percent of CO2, we need to find the moles of CO and CO2 in the mixture. Since the moles of CO and CO2 remain the same before and after the reaction, the mole percent of CO2 in the mixture is given by:
Mole percent of CO2 = (moles of CO2 / total moles) * 100
Since the total moles remain the same, the mole percent of CO2 will also remain the same. Therefore, the mole percent of CO2 in the mixture is indeed 60.
c) Mole fraction of CO in the mixture is 0.40:
To calculate the mole fraction of CO, we need to find the moles of CO and CO2 in the mixture. Since the moles of CO and CO2 remain the same before and after the reaction, the mole fraction of CO is given by:
Mole fraction of CO = moles of CO / total moles
Since the total moles remain the same, the mole fraction of CO will remain the same. Therefore, the mole fraction of CO in the mixture is indeed 0.40.
d) The mixture contains 40 mL of CO:
Since the total volume of the mixture increased from 100 mL to 160 mL, it implies that the reaction consumed some of the CO gas. Therefore, the mixture does not contain 40 mL of CO.
Conclusion:
Based on the analysis, the correct statements are:
- The mixture contains 40 mL of CO2
- The mole percent of CO2 in the mixture is 60
- The mole fraction of CO in the mixture is 0.40
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Community Answer
A 100 mL mixture of CO and CO2 is passed through a tube containing re...
CO2 (g) + C(s) ⟶ 2 CO (g)
Vol of CO2 = x mL
Vol of CO = (100 - x) mL
Final volume = (100 - x + 2x) mL
= 100 + x = 160 ∴ x = 60 mL
Volume of CO = 40 mL
Mole-fraction of CO = 40 / 100 = 0.4
Mol % CO2 = 60%
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A 100 mL mixture of CO and CO2 is passed through a tube containing red hot charcoal. The volume now becomes 160 mL. The volumes are measured under the same conditions of temperature and pressure. Amongst the following, select the correct statement (s).a)The mixture contains 40 mL of CO2b)Mole percent of CO2 in the mixture is 60.c)Mole fraction of CO in the mixture is 0.40d)The mixture contains 40 mL of COCorrect answer is option 'B,C,D'. Can you explain this answer?
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A 100 mL mixture of CO and CO2 is passed through a tube containing red hot charcoal. The volume now becomes 160 mL. The volumes are measured under the same conditions of temperature and pressure. Amongst the following, select the correct statement (s).a)The mixture contains 40 mL of CO2b)Mole percent of CO2 in the mixture is 60.c)Mole fraction of CO in the mixture is 0.40d)The mixture contains 40 mL of COCorrect answer is option 'B,C,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 A 100 mL mixture of CO and CO2 is passed through a tube containing red hot charcoal. The volume now becomes 160 mL. The volumes are measured under the same conditions of temperature and pressure. Amongst the following, select the correct statement (s).a)The mixture contains 40 mL of CO2b)Mole percent of CO2 in the mixture is 60.c)Mole fraction of CO in the mixture is 0.40d)The mixture contains 40 mL of COCorrect answer is option 'B,C,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 A 100 mL mixture of CO and CO2 is passed through a tube containing red hot charcoal. The volume now becomes 160 mL. The volumes are measured under the same conditions of temperature and pressure. Amongst the following, select the correct statement (s).a)The mixture contains 40 mL of CO2b)Mole percent of CO2 in the mixture is 60.c)Mole fraction of CO in the mixture is 0.40d)The mixture contains 40 mL of COCorrect answer is option 'B,C,D'. Can you explain this answer?.
A 100 mL mixture of CO and CO2 is passed through a tube containing red hot charcoal. The volume now becomes 160 mL. The volumes are measured under the same conditions of temperature and pressure. Amongst the following, select the correct statement (s).a)The mixture contains 40 mL of CO2b)Mole percent of CO2 in the mixture is 60.c)Mole fraction of CO in the mixture is 0.40d)The mixture contains 40 mL of COCorrect answer is option 'B,C,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 A 100 mL mixture of CO and CO2 is passed through a tube containing red hot charcoal. The volume now becomes 160 mL. The volumes are measured under the same conditions of temperature and pressure. Amongst the following, select the correct statement (s).a)The mixture contains 40 mL of CO2b)Mole percent of CO2 in the mixture is 60.c)Mole fraction of CO in the mixture is 0.40d)The mixture contains 40 mL of COCorrect answer is option 'B,C,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 A 100 mL mixture of CO and CO2 is passed through a tube containing red hot charcoal. The volume now becomes 160 mL. The volumes are measured under the same conditions of temperature and pressure. Amongst the following, select the correct statement (s).a)The mixture contains 40 mL of CO2b)Mole percent of CO2 in the mixture is 60.c)Mole fraction of CO in the mixture is 0.40d)The mixture contains 40 mL of COCorrect answer is option 'B,C,D'. Can you explain this answer?.
Solutions for A 100 mL mixture of CO and CO2 is passed through a tube containing red hot charcoal. The volume now becomes 160 mL. The volumes are measured under the same conditions of temperature and pressure. Amongst the following, select the correct statement (s).a)The mixture contains 40 mL of CO2b)Mole percent of CO2 in the mixture is 60.c)Mole fraction of CO in the mixture is 0.40d)The mixture contains 40 mL of COCorrect answer is option 'B,C,D'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
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A 100 mL mixture of CO and CO2 is passed through a tube containing red hot charcoal. The volume now becomes 160 mL. The volumes are measured under the same conditions of temperature and pressure. Amongst the following, select the correct statement (s).a)The mixture contains 40 mL of CO2b)Mole percent of CO2 in the mixture is 60.c)Mole fraction of CO in the mixture is 0.40d)The mixture contains 40 mL of COCorrect answer is option 'B,C,D'. Can you explain this answer?, a detailed solution for A 100 mL mixture of CO and CO2 is passed through a tube containing red hot charcoal. The volume now becomes 160 mL. The volumes are measured under the same conditions of temperature and pressure. Amongst the following, select the correct statement (s).a)The mixture contains 40 mL of CO2b)Mole percent of CO2 in the mixture is 60.c)Mole fraction of CO in the mixture is 0.40d)The mixture contains 40 mL of COCorrect answer is option 'B,C,D'. Can you explain this answer? has been provided alongside types of A 100 mL mixture of CO and CO2 is passed through a tube containing red hot charcoal. The volume now becomes 160 mL. The volumes are measured under the same conditions of temperature and pressure. Amongst the following, select the correct statement (s).a)The mixture contains 40 mL of CO2b)Mole percent of CO2 in the mixture is 60.c)Mole fraction of CO in the mixture is 0.40d)The mixture contains 40 mL of COCorrect answer is option 'B,C,D'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A 100 mL mixture of CO and CO2 is passed through a tube containing red hot charcoal. The volume now becomes 160 mL. The volumes are measured under the same conditions of temperature and pressure. Amongst the following, select the correct statement (s).a)The mixture contains 40 mL of CO2b)Mole percent of CO2 in the mixture is 60.c)Mole fraction of CO in the mixture is 0.40d)The mixture contains 40 mL of COCorrect answer is option 'B,C,D'. Can you explain this answer? tests, examples and also practice JEE tests.
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