What is the mass fraction of C2HCl3 in the liquid phase? 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?
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.