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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 option '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
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The vapour density of a mixture of gas A (Molecular mass = 40) and gas...
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The vapour density of a mixture of gas A (Molecular mass = 40) and gas...
Given:
Molecular mass of gas A = 40
Molecular mass of gas B = 80
Vapour density of the mixture = 25

To find: Mole % of gas B in the mixture

Formula:
Vapour density = (Molecular mass of gas A x mole % of gas A + Molecular mass of gas B x mole % of gas B) / 100

Calculation:
Substituting the given values in the above formula, we get:
25 = (40 x mole % of gas A + 80 x mole % of gas B) / 100

Simplifying the above equation, we get:
10 = mole % of gas A + 2 x mole % of gas B

We know,
mole % of gas A + mole % of gas B = 100

Substituting the above equation in the equation obtained earlier, we get:
10 = 100 - mole % of gas B + 2 x mole % of gas B

Solving for mole % of gas B, we get:
Mole % of gas B = 25%

Therefore, the correct option is (a) 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 option 'A'. Can you explain this answer?
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