Question

If a sample of 12.0 grams of fluorine gas at 45.00C has a pressure of 0.544 atm, what is the volume of the container?

R = 0.0821 L atm/mol K



15.1 L


4.29 L


30.3 L


2.14 L

Answer 1

Answer: Volume of the container is 30.3 L.

Step-by-step explanation:

Given: Mass = 12 g

Temperature =
`45^(o)C` = (45 + 273) K = 318 K

Pressure = 0.544 atm

R = gas constant = 0.0821 L atm/mol K

Moles is the mass of substance divided by its molar mass.

Hence, moles of fluorine (molar mass = 18.99 g/mol) are as follows.

`Moles = (mass)/(molar mass)\\= (12.0 g)/(18.99 g/mol)\\= 0.631 mol`

Formula used to calculate the volume of container is as follows.

PV = nRT

where,

P = pressure

V = volume

n = no. of moles

R = gas constant

T = temperature

Substitute the value into above formula as follows.

`PV = nRT\\0.544 atm * V = 0.631 mol * 0.0821 L atm/mol K * 318 K\\V = 30.3 L`

Thus, we can conclude that volume of the container is 30.3 L.