Respuesta :
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
Based on the mass of gas obtained and initial mass of CCl3F in the container, 5.19 g of CCl3F will be in liquid form.
What is the mass of gas present?
The mass of gas present is found first by determining the moles of gas present using the ideal gas equation.
Then the mass of gas is determined using the formula:
- mass = moles * molar mass
Using the ideal gas equation:
- PV = nRT
Where:
- P: Pressure
- V: Volume
- n: number of moles
- R: gas constant
- T: Temperature
making n subject of formula:
- n = PV/RT
From data provided:
P = 856/760 = 1.13 atm
R = 0.082 L atm / mol K
then:
n = 1.13 * 1 / 0.082 * 300
n = 0.0459 moles
Now, the mass of gas is determined:
molar mass of CCl3F = 137.37 g/mol so:
mass of gas = 0.0459 * 137.37
mass of gas = 6.31 g
What of CCL3F is liquid?
The mass of the liquid CCL3F is obtained from the formula:
- mass of liquid = total mass - mass of gas
mass of liquid = 11.5 - 6.31
mass of liquid = 5.19 g
Therefore, 5.19 g of CCL3F will be in liquid form.
Learn more about vapour pressure and mass at: https://brainly.com/question/4373476
