Part A) Compounds Q and R are mixed at 298, creating an ideal solution. If the m

Question

Part A)

Compounds Q and R are mixed at 298, creating an ideal solution. If the mole fraction of R in the liquid phase is 0.68, and the pure vapor pressures of Q and R are 84 Torr and 40 Torr, respectively, what are the partial pressures of Q and R in the gas phase?

**I know that the partial pressure of Q is 26.88 torr and the partial pressure of R is 27.2 torr. I am having troubles answering part B given these answers to part A. Please only answer part B given these answers for part A. ***

Part B)

If I remove some gas from the container in question 4 and condense it in a new container, what are the partial pressures of Q and R after a new equilibrium is established at 298 K in the new container?

Explanation / Answer

part B)

When the vapours from container of part A are taken into new container and condensed to liquid.

Thus the molefractions of Q and R in vapor state in the previous condenser are now the molefractions in liquid of new condenser.(we condensed the vapors of previous condenser).

Let us calculate the mole fractions of Q and R in vapor in part A.

The partial pressure of Q = 26.88torr

and partial pressure of R = 27.2 torr

So total vapor pressure = Pq +Pr

= 26.88+ 27.2

= 54.08 torr

The mole fraction of Q in vapor = partial pressure of Q / total pressure [Dalton's law]

= 26.88/54.08

=0.497

thus mole fraction of R in vapor = 1-0.497

=0.503

Now this vapor is condensed in to liquid.

So the container in part B has a solution with mole fraction of Q =0.497 and mole fraction of R =0.503

So partial pressure of Q in new container = mole fraction x Pressure of pure solvent Q [Raoult's law]

= 0.497x84 torr

=41.748 torr

and partial pressure of R in new container = 0.503 x40 torr

= 20.12 Torr

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