If 11.0 g of ccl3f is enclosed in a 1.1 −l container, will any liquid be present? if so, what mass of liquid?

Considering that CCL3F gas behave like an ideal gas then we can use the Ideal Gas Law 
PV = nRT, however is an approximation and not the only way to resolve this problem with the given data..So,at the end of the solution I am posting some sources for further understanding and a expanded point of view. 

Data: P= 856torr, T = 300K, V= 1.1L, R = 62.36 L Torr / KMol 

Solving and substituting in the Gas equation for n = PV / RT = (856)(1.1L) /( 62.36)(300) = 0.05 Mol. This RESULT is of any gas. To tie it up to our gas we need to look for its molecular weight:MW of CCL3F = 137.7 gm/mol.  

Then : 0.05x 137.5 = 6.88gm of vapor 

If we sustract the vapor weight from the TOTAL weight of liquid we have: 11.5gm - 6.88gm = 4.62 gm of liquid.d

I’d have to say no because without oxygen in the water the fish would all be dead. so if you want to fish for dead fish, pick the ponds with the lowest oxygen concentrations.