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The vapour density of a mixture of gas A (Molecular mass = 40) and gas B (Molecular mass = 80) is 25.Then mole % of gas B in the mixture would be
- a)25%
- b)50%
- c)75%
- d)10%
Correct answer is 'A'. Can you explain this answer?
Verified Answer
The vapour density of a mixture of gas A (Molecular mass = 40) and gas...
Let mole of B = x
V.D = 25 mole of A = 100 x
Mol. mass = 50
V.D = 25 mole of A = 100 x
Mol. mass = 50
This question is part of UPSC exam. View all Class 11 courses
Most Upvoted Answer
The vapour density of a mixture of gas A (Molecular mass = 40) and gas...
Calculation of Mole Fraction of Gas B in the Mixture
Given, vapour density of the mixture = 25
We know that,
Vapour Density = (Molecular Weight of Mixture) / 2
Therefore,
25 = (Molecular Weight of Mixture) / 2
Molecular Weight of Mixture = 50
Now, let the mole fraction of gas A in the mixture be x, and that of gas B be (1-x).
Molecular Weight of Mixture = (Molecular Weight of A) * x + (Molecular Weight of B) * (1-x)
50 = 40x + 80(1-x)
50 = 40x + 80 - 80x
30x = 30
x = 1
Therefore, mole fraction of gas A in the mixture = 1
Mole fraction of gas B in the mixture = 1-1 = 0
Hence, the mole % of gas B in the mixture would be 0, which is equivalent to 0% and the correct answer is option A.
Given, vapour density of the mixture = 25
We know that,
Vapour Density = (Molecular Weight of Mixture) / 2
Therefore,
25 = (Molecular Weight of Mixture) / 2
Molecular Weight of Mixture = 50
Now, let the mole fraction of gas A in the mixture be x, and that of gas B be (1-x).
Molecular Weight of Mixture = (Molecular Weight of A) * x + (Molecular Weight of B) * (1-x)
50 = 40x + 80(1-x)
50 = 40x + 80 - 80x
30x = 30
x = 1
Therefore, mole fraction of gas A in the mixture = 1
Mole fraction of gas B in the mixture = 1-1 = 0
Hence, the mole % of gas B in the mixture would be 0, which is equivalent to 0% and the correct answer is option A.
Community Answer
The vapour density of a mixture of gas A (Molecular mass = 40) and gas...
Vapour density= molecular mass/2
i.e., average molecular mass of the mixture=50
let us assume no. of moles of A=x and B=y
50=40x+80y/x+y
50x+50y=40x+80y
10x=30y
x=3y
%age of B
1/4*100=25%
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The vapour density of a mixture of gas A (Molecular mass = 40) and gas B (Molecular mass = 80) is 25.Then mole % of gas B in the mixture would bea)25%b)50%c)75%d)10%Correct answer is 'A'. Can you explain this answer?
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The vapour density of a mixture of gas A (Molecular mass = 40) and gas B (Molecular mass = 80) is 25.Then mole % of gas B in the mixture would bea)25%b)50%c)75%d)10%Correct answer is '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 The vapour density of a mixture of gas A (Molecular mass = 40) and gas B (Molecular mass = 80) is 25.Then mole % of gas B in the mixture would bea)25%b)50%c)75%d)10%Correct answer is '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 The vapour density of a mixture of gas A (Molecular mass = 40) and gas B (Molecular mass = 80) is 25.Then mole % of gas B in the mixture would bea)25%b)50%c)75%d)10%Correct answer is 'A'. Can you explain this answer?.
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