Answer:
The mass of fluorine gas sample in the stainless steel container is
522.455 grams
Explanation:
To solve the question, we note the given parameters to see if we have enough facts to provide an answer
Volume of container = 31 L = 0.031 m³
Pressure of the fluorine in the container = 160 psi
Temperature of the fluorine gas in the container = 26 °C
We have the universal gas equation given by
P·V = n·R·T
Where P = Pressure 160 psi = 1103161 Pa
V = Volume = 31 L = 0.031 m³
n = Number of moles
T = Temperature = 26 °C = (273.15 + 26) K= 299.15 K and
R = Universal gas constant = 8.314 J/mol·K
Therefore n = [tex]\frac{PV}{RT}[/tex] = [tex]\frac{1103161*0.031}{8.314*299.15}[/tex] = 13.75 moles of fluorine
The molar mass of fluorine gas = 37.99681 g/mol
Therefore the mass of fluorine gas contained in the stainless steel container is 13.75 moles × 37.99681 g/mol = 522.455 grams
