3A. The vapor pressure of pure liquid A is 575 Torr and of pure liquid B is 390

Question

3A. The vapor pressure of pure liquid A is 575 Torr and of pure liquid B is 390 Torr at a temperature of 300 K. The liquids are combined to form a solution. The mole fraction of the liquid phase is 0.40 in A. Calculate the equilibrium partial pressures of A and B and the total pressure above the solution. Assume ideal solution behavior. 3B. Continuation of 3. Calculate the mole fraction of components A and B in the vapor/gas phase at equilibrium over the solution. Which component is enriched in the vapor phase relative to the liquid phase? Which component is more volatile? 3C. (1 pt) Continuation of 3. Suppose you were distilling a mixture of A and B. Which solution would be in the receiving flask?

Explanation / Answer

Partial pressure = vapour pressure of pure liquid multiplied by its mole fraction that is , p1=p10 * x1

For A , p10 = 575 torr and  x1 = 0.4 so partial pressure of A = 575 * 0.4 = 230 torr

For B, p10 = 390 torr and  x1 = 0.6 (as 1-xa) so partial pressure of B = 390 * 0.6 = 234 torr

total pressure = 230 + 234 = 464 torr

3B. partial pressure = mole fraction in vapour phase * total pressure

mole fraction of A in vapour phase = 230/ 464= 0.496

mole fraction of B = 1 - mole fraction of A = 1 - 0.496 = 0.504

so, B is enriched in vapour phase.

Component is volatile if it has high pressure in pure form , so A is more volatile.

3C. As A is more volatile, receiving flask will have A.

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