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At 300 K two pure liquids A and B have vapour pressures respectively 150 mm Hg and 100 mm Hg. In an equimolar liquid mixture of A and B, the mole fraction of B in the vapour mixture at this temp. is :
  • a)
    0.6
  • b)
    0.5
  • c)
    0.8
  • d)
    0.4
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
At 300 K two pure liquids A and B have vapour pressures respectively 1...
Let the moles of A = x
∴ The moles of B = x (∵ mixture is equimolar)
Mole fraction of A = 
= Mole fraction of B
Thus, 
ρT = 150 x 0.5 + 100 x 0.5 = 125
∴ Mole fraction of B in the vapour mixture = 

Hence, option D is correct.
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At 300 K two pure liquids A and B have vapour pressures respectively 150 mm Hg and 100 mm Hg. In an equimolar liquid mixture of A and B, the mole fraction of B in the vapour mixture at this temp. is :a)0.6b)0.5c)0.8d)0.4Correct answer is option 'D'. Can you explain this answer?
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