Suppose a $250.mL$ flask is filled with $2\mathrm{.}0$mol of $C{l}_2,0\mathrm{.}70$mol of $CHC{l}_3$ and $0\mathrm{.}40$mol of $HCl.$ The following reaction becomes possible:

$C{l}_2(g)+CHC{l}_3(g)HCl(g)+CC{l}_4(g)$

The equilibrium constant $K$ for this reaction is $8\mathrm{.}79$ at the temperature of the flask.

Calculate the equilibrium molarity of $CHC{l}_3.$ Round your answer to two decimal places.