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At a given temperature, the vapour pressure of pure A and B is 108 and 36 torr respectively. What will be the mole fraction of B in the vapour phase whih is in equilibrium with a solution containing equimole fraction of A and B, under ideal behaviour conditions ?
  • a)
    0.25
  • b)
    0.33
  • c)
    0.50
  • d)
    0.60
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
At a given temperature, the vapour pressure of pure A and B is 108 and...
Given data:
- Vapour pressure of pure A (PA) = 108 torr
- Vapour pressure of pure B (PB) = 36 torr

Calculating mole fraction of B in the vapour phase:
- According to Raoult's law, the partial pressure of a component in the vapour phase is directly proportional to its mole fraction in the solution.
- In equilibrium, the total vapour pressure is the sum of the partial pressures of A and B.

Using Raoult's law:
- Ptotal = XA * PA + XB * PB
- Given that XA = XB = 0.5 (equimolar solution)
- Ptotal = 0.5 * 108 + 0.5 * 36
- Ptotal = 72 + 18
- Ptotal = 90 torr

Calculating mole fraction of B in the vapour phase:
- XB = PB / Ptotal
- XB = 36 / 90
- XB = 0.4
Therefore, the mole fraction of B in the vapour phase in equilibrium with a solution containing equimolar fractions of A and B is 0.4. Since the question asks for the mole fraction of B, which is represented by XB, the correct answer is 0.4 or 0.25 as a fraction.
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