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Two liquids X and Y form an ideal solution. At 300 K, vapour pressure of the solution containing 1 mol of X and 3 mol of Y is 550 mm Hg. At the same temperature, if 1 mol of Y is further added to this solution, vapour pressure of the solution increases by 10 mm Hg. Vapour pressure (in mm Hg) of X and Y in their pure states will be, respectively.
  • a)
    200 and 300
  • b)
    300 and 400
  • c)
    400 and 600
  • d)
    500 and 600
Correct answer is option 'C'. Can you explain this answer?
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Two liquids X and Y form an ideal solution. At 300 K, vapour pressure ...
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Two liquids X and Y form an ideal solution. At 300 K, vapour pressure ...
Given Data
- Vapour pressure of the solution (1 mol X + 3 mol Y) = 550 mm Hg
- Vapour pressure increase upon adding 1 mol of Y = 10 mm Hg
Ideal Solution Concept
In an ideal solution, the vapour pressure is calculated using Raoult's Law:
P_solution = X_X * P°_X + X_Y * P°_Y
Where:
- P_solution = vapour pressure of the solution
- X_X = mole fraction of X
- X_Y = mole fraction of Y
- P°_X = vapour pressure of pure X
- P°_Y = vapour pressure of pure Y
Mole Fractions Calculation
Initially, for the solution of 1 mol of X and 3 mol of Y:
- Total moles = 1 + 3 = 4
- Mole fraction of X (X_X) = 1/4 = 0.25
- Mole fraction of Y (X_Y) = 3/4 = 0.75
Substituting into Raoult's law:
550 = 0.25 * P°_X + 0.75 * P°_Y
After Adding 1 Mol of Y
New total moles = 1 + 4 = 5
- New mole fraction of X = 1/5 = 0.2
- New mole fraction of Y = 4/5 = 0.8
Using the increase in vapour pressure:
560 = 0.2 * P°_X + 0.8 * P°_Y
Solving the Equations
Now, we have two equations:
1. 550 = 0.25 * P°_X + 0.75 * P°_Y
2. 560 = 0.2 * P°_X + 0.8 * P°_Y
Solving these equations simultaneously yields:
P°_X = 400 mm Hg and P°_Y = 600 mm Hg.
Conclusion
Thus, the vapour pressures of pure X and Y are:
- Pure X: 400 mm Hg
- Pure Y: 600 mm Hg
The correct answer is option 'C'.
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Two liquids X and Y form an ideal solution. At 300 K, vapour pressure of the solution containing 1 mol of X and 3 mol of Y is 550 mm Hg. At the same temperature, if 1 mol of Y is further added to this solution, vapour pressure of the solution increases by 10 mm Hg. Vapour pressure (in mm Hg) of X and Y in their pure states will be, respectively.a) 200 and 300 b) 300 and 400 c) 400 and 600 d) 500 and 600 Correct answer is option 'C'. Can you explain this answer?
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