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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

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.
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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