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Vapour-liquid equilibrium of a two-component ideal solution of trichloroethene (C2HCl3) and trichloromethane (CHCl3) is established at 25 °C. The mole fraction of CHCl3 in the vapour phase is 1. What is the mass fraction of C2HCl3 in the liquid phase? The vapor pressure of trichloroethene and trichloromethane at 25 °C are:

Pvap, C2HCl3 = 73.0 mmHg

Pvap, CHCl3 = 199.1 mm Hg

I'm confused about this problem, if the mole fraction of one element equals to 1, would not be the mole fraction of another element be 0? Can someone advise me on how to approach this problem?

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  • $\begingroup$ "I'm confused about this problem, if the mole fraction of one element equals to 1, would not be the mole fraction of another element be 0?" // Yes. $\endgroup$
    – MaxW
    Nov 8, 2015 at 5:27
  • $\begingroup$ The system is supposed to be "ideal." What does that mean to mixture in vapor vs mixture in liquid? // Hint you need the boiling point of the two compounds. $\endgroup$
    – MaxW
    Nov 8, 2015 at 5:28
  • $\begingroup$ @MaxW but i don't get it, if the mole fraction of another compound is 0, wouldn't that mean that the moles of this compound is 0? The only way to get mole fraction to be 0, is to have 0 in the numerator which is the number of moles of the compound. $\endgroup$
    – Jack
    Nov 8, 2015 at 5:44
  • $\begingroup$ Yes that is correct. What happens when the temperature changes for this system? $\endgroup$
    – MaxW
    Nov 8, 2015 at 5:48
  • $\begingroup$ @MaxW so, at boiling temperatures the total pressure of the system will be the same as the partial pressure of the CHCL3? And then, from here should i use PV = nRT to find the moles of CHCL3 in order to find the mass fraction of C2HCL3. But then, i don't know the volume and if the number of moles of C2HCL3 is zero, wouldn't that mean that the mass fraction of it is zero? $\endgroup$
    – Jack
    Nov 8, 2015 at 5:54

1 Answer 1

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I'm sorry. I mislead you. Temperature changes, and the boiling points of the liquid have nothing to do with this problem as I erroneously commented.

The correct approach is to simply apply Raoult's law. For some temperature T, let

$p^*_i =$ partial pressure of some pure compound $i$
$x_i =$ mole fraction of the compound $i$ in solution.

then let $p_i$ be the partial pressure of compound i above a liquid mixture. For an ideal system then:

$p_i = p^*_ix_i$

The problem states "The mole fraction of $\ce{CHCl3}$ in the vapor phase is 1."

As you deduced, since there is no $\ce{C2HCl3}$ in the vapor phase there can't be any $\ce{C2HCl3}$ in the liquid phase either.

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