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

Question

asked Sep 1, 2022 90.6k views

1 Answer

Karln · Answer 1 · 2022-09-08T12:00:14+0000

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.