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