Question

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?

Answer 1

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

Answer 2

Yes, liquid is present and the quantity of liquid present is; 4.81g of ccl3f liquid

At standard temperature and pressure;

• Pressure = 1 atm and Temp. = 298K.
• Gas constant = 0.0821 L-atm/Kmol.

By the ideal gas equation; the no. of moles of ccl3f gas present in the 1.1 L container is;

• n = PV/RT

• n = (1 × 1.1)/(0.0821 × 298)

• n = 0.045 moles of gas.

Therefore, the mass of gas present can be calculated as;

• Mass of gas = 0.045 × 137.5

• where; 137.5 is the molar mass of ccl3f.

Mass of gas = 6.19 grams of CCl3F.

Ultimately, liquid is present and the mass of liquid present is; 11g - 6.19g = 4.81g of ccl3f liquid.

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