Question

The vapor pressure of CCl3F at 300 K is 856 torr. If 11.5 g of CCl F is enclosed in a 1.0-L container, will any liquid be pres- vap Cs,solid 118 J>mol 3 ent? If so, what mass of liquid?

Answer 1

Answer:

Yes, there will be liquid present and the mass is 5.19 g

Explanation:

In order to do this, we need to use the equation of an ideal gas which is:

PV = nRT (1)

Where:

P: Pressure

V: Volume

n: number of moles

R: gas constant

T: Temperature

we know that the pressure is 856 Torr at 300 K. So, if we want to know if there'll be any liquid present, we need to calculate the moles and mass of the CCl3F at this pressure and temperature, and then, compare it to the initial mass of 11.5 g.

From (1), solving for moles we have:

n = PV/RT (2)

Solving for n:

P = 856/760 = 1.13 atm

R = 0.082 L atm / mol K

n = 1.13 * 1 / 0.082 * 300

n = 0.0459 moles

Now, the mass is:

m = n * MM (3)

The molar mass of CCl3F reported is 137.37 g/mol so:

m = 0.0459 * 137.37

m = 6.31 g

Finally, this means that if we put 11.5 g of CCl3F in a container, only 6.31 g will become gaseous, so, this means it will be liquid present, and the mass is:

m = 11.5 - 6.31

m = 5.19 g