Class 11 Exam > Class 11 Questions > A and B are two identical vessels. A contains...
Start Learning for Free
A and B are two identical vessels. A contains 7.5 g ethane at 1atm and 298K. The vessel B contains 37.5 g of a gas X at same temperature and pressure. The vapour density of X is :
- a)75
- b)150
- c)37.5
- d)45
Correct answer is option 'A'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Verified Answer
A and B are two identical vessels. A contains 7.5 g ethane at 1atm and...
Most Upvoted Answer
A and B are two identical vessels. A contains 7.5 g ethane at 1atm and...
Given:
- Two identical vessels A and B.
- Vessel A contains 15 g of ethane (C2H6) at 1 atm and 298 K.
- Vessel B contains 75 g of a gas X2 at the same temperature and pressure.
To find:
The vapor density of gas X2.
Solution:
1. Understanding Vapor Density:
- Vapor density is the ratio of the molar mass of a gas to the molar mass of hydrogen gas.
- It is a measure of how heavy a gas is compared to hydrogen gas.
- The vapor density of a gas can be calculated using the formula:
Vapor density = (Molar mass of gas) / (Molar mass of hydrogen gas)
2. Calculating the Molar Mass of Ethane:
- Ethane (C2H6) is a hydrocarbon composed of 2 carbon atoms (C) and 6 hydrogen atoms (H).
- The molar mass of carbon (C) is 12.01 g/mol, and the molar mass of hydrogen (H) is 1.01 g/mol.
- Therefore, the molar mass of ethane (C2H6) can be calculated as:
Molar mass of ethane = (2 * Molar mass of carbon) + (6 * Molar mass of hydrogen)
= (2 * 12.01 g/mol) + (6 * 1.01 g/mol)
= 24.02 g/mol + 6.06 g/mol
= 30.08 g/mol
3. Calculating the Molar Mass of Gas X2:
- The molar mass of gas X2 can be calculated using the given mass of gas X2.
- Vessel B contains 75 g of gas X2.
- The molar mass of gas X2 can be calculated using the formula:
Molar mass of gas X2 = (Mass of gas X2) / (Number of moles of gas X2)
- To find the number of moles of gas X2, we need to use the ideal gas equation:
PV = nRT
Where:
P = Pressure of the gas (1 atm)
V = Volume of the gas (unknown)
n = Number of moles of the gas (unknown)
R = Ideal gas constant (0.0821 L.atm/mol.K)
T = Temperature of the gas (298 K)
- Rearranging the ideal gas equation to solve for the number of moles (n):
n = (PV) / (RT)
- Since the volume (V) is the same for both vessels A and B, the number of moles of gas X2 in vessel B is the same as the number of moles of ethane in vessel A.
- Using the number of moles of ethane in vessel A, we can calculate the molar mass of gas X2:
Molar mass of gas X2 = (Mass of gas X2) / (Number of moles of gas X2)
= 75 g / (n moles)
= 75 g / (n moles of ethane in vessel A)
4. Comparing the Molar Mass of Ethane and Gas X2
- Two identical vessels A and B.
- Vessel A contains 15 g of ethane (C2H6) at 1 atm and 298 K.
- Vessel B contains 75 g of a gas X2 at the same temperature and pressure.
To find:
The vapor density of gas X2.
Solution:
1. Understanding Vapor Density:
- Vapor density is the ratio of the molar mass of a gas to the molar mass of hydrogen gas.
- It is a measure of how heavy a gas is compared to hydrogen gas.
- The vapor density of a gas can be calculated using the formula:
Vapor density = (Molar mass of gas) / (Molar mass of hydrogen gas)
2. Calculating the Molar Mass of Ethane:
- Ethane (C2H6) is a hydrocarbon composed of 2 carbon atoms (C) and 6 hydrogen atoms (H).
- The molar mass of carbon (C) is 12.01 g/mol, and the molar mass of hydrogen (H) is 1.01 g/mol.
- Therefore, the molar mass of ethane (C2H6) can be calculated as:
Molar mass of ethane = (2 * Molar mass of carbon) + (6 * Molar mass of hydrogen)
= (2 * 12.01 g/mol) + (6 * 1.01 g/mol)
= 24.02 g/mol + 6.06 g/mol
= 30.08 g/mol
3. Calculating the Molar Mass of Gas X2:
- The molar mass of gas X2 can be calculated using the given mass of gas X2.
- Vessel B contains 75 g of gas X2.
- The molar mass of gas X2 can be calculated using the formula:
Molar mass of gas X2 = (Mass of gas X2) / (Number of moles of gas X2)
- To find the number of moles of gas X2, we need to use the ideal gas equation:
PV = nRT
Where:
P = Pressure of the gas (1 atm)
V = Volume of the gas (unknown)
n = Number of moles of the gas (unknown)
R = Ideal gas constant (0.0821 L.atm/mol.K)
T = Temperature of the gas (298 K)
- Rearranging the ideal gas equation to solve for the number of moles (n):
n = (PV) / (RT)
- Since the volume (V) is the same for both vessels A and B, the number of moles of gas X2 in vessel B is the same as the number of moles of ethane in vessel A.
- Using the number of moles of ethane in vessel A, we can calculate the molar mass of gas X2:
Molar mass of gas X2 = (Mass of gas X2) / (Number of moles of gas X2)
= 75 g / (n moles)
= 75 g / (n moles of ethane in vessel A)
4. Comparing the Molar Mass of Ethane and Gas X2
Attention Class 11 Students!
To make sure you are not studying endlessly, EduRev has designed Class 11 study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in Class 11.
|
Explore Courses for Class 11 exam
|
|
Similar Class 11 Doubts
Top Courses for Class 11View all
A and B are two identical vessels. A contains 7.5 g ethane at 1atm and 298K. The vessel B contains 37.5 g of a gas Xat same temperature and pressure. The vapour density of X is :a)75b)150c)37.5d)45Correct answer is option 'A'. Can you explain this answer?
Question Description
A and B are two identical vessels. A contains 7.5 g ethane at 1atm and 298K. The vessel B contains 37.5 g of a gas Xat same temperature and pressure. The vapour density of X is :a)75b)150c)37.5d)45Correct answer is option 'A'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A and B are two identical vessels. A contains 7.5 g ethane at 1atm and 298K. The vessel B contains 37.5 g of a gas Xat same temperature and pressure. The vapour density of X is :a)75b)150c)37.5d)45Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A and B are two identical vessels. A contains 7.5 g ethane at 1atm and 298K. The vessel B contains 37.5 g of a gas Xat same temperature and pressure. The vapour density of X is :a)75b)150c)37.5d)45Correct answer is option 'A'. Can you explain this answer?.
A and B are two identical vessels. A contains 7.5 g ethane at 1atm and 298K. The vessel B contains 37.5 g of a gas Xat same temperature and pressure. The vapour density of X is :a)75b)150c)37.5d)45Correct answer is option 'A'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A and B are two identical vessels. A contains 7.5 g ethane at 1atm and 298K. The vessel B contains 37.5 g of a gas Xat same temperature and pressure. The vapour density of X is :a)75b)150c)37.5d)45Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A and B are two identical vessels. A contains 7.5 g ethane at 1atm and 298K. The vessel B contains 37.5 g of a gas Xat same temperature and pressure. The vapour density of X is :a)75b)150c)37.5d)45Correct answer is option 'A'. Can you explain this answer?.
Solutions for A and B are two identical vessels. A contains 7.5 g ethane at 1atm and 298K. The vessel B contains 37.5 g of a gas Xat same temperature and pressure. The vapour density of X is :a)75b)150c)37.5d)45Correct answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for Class 11.
Download more important topics, notes, lectures and mock test series for Class 11 Exam by signing up for free.
Here you can find the meaning of A and B are two identical vessels. A contains 7.5 g ethane at 1atm and 298K. The vessel B contains 37.5 g of a gas Xat same temperature and pressure. The vapour density of X is :a)75b)150c)37.5d)45Correct answer is option 'A'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
A and B are two identical vessels. A contains 7.5 g ethane at 1atm and 298K. The vessel B contains 37.5 g of a gas Xat same temperature and pressure. The vapour density of X is :a)75b)150c)37.5d)45Correct answer is option 'A'. Can you explain this answer?, a detailed solution for A and B are two identical vessels. A contains 7.5 g ethane at 1atm and 298K. The vessel B contains 37.5 g of a gas Xat same temperature and pressure. The vapour density of X is :a)75b)150c)37.5d)45Correct answer is option 'A'. Can you explain this answer? has been provided alongside types of A and B are two identical vessels. A contains 7.5 g ethane at 1atm and 298K. The vessel B contains 37.5 g of a gas Xat same temperature and pressure. The vapour density of X is :a)75b)150c)37.5d)45Correct answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A and B are two identical vessels. A contains 7.5 g ethane at 1atm and 298K. The vessel B contains 37.5 g of a gas Xat same temperature and pressure. The vapour density of X is :a)75b)150c)37.5d)45Correct answer is option 'A'. Can you explain this answer? tests, examples and also practice Class 11 tests.
|
|
Explore Courses for Class 11 exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup