If 11.5 g of CCl3F is placed in a 1.2 −L container, is any liquid present?
Neglect the volume of liquid, if any, in the vial:
n = PV / RT = (856 torr) x (1.2 L) / ((62.36367 L torr/K mol) x (300 K)) = 0.054904 mol of gas
(0.054904 mol CCl3F) x (137.3681 g CCl3F/mol) = 7.54 g CCl3F in gaseous state
So, yes, there will be some liquid left.
(11.5 g total) – (7.54 g gas) = 3.96 g liquid