Answer:
The mass of oxygen the container must have is 150.85 g.
Explanation:
Given;
mass of the oxygen, m₁ = 44.5 g
initial pressure of the gas, P₁ = 2.3 atm
final pressure of the gas, P₂ = 7.8 atm
Atomic mass of oxygen gas, = O₂ = 16 x 2 = 32 g
initial number of moles of oxygen in the container, n₁ = 44.5/32 = 1.39
let the final number of moles of oxygen = n₂
Apply ideal gas equation;
PV = nRT
[tex]\frac{PV}{Rn} = T\\\\since \ temperature\ T \ is \ constant;\\\\\frac{P_1V}{Rn_1} = \frac{P_2V}{Rn_2}\\\\\frac{P_1}{n_1} = \frac{P_2}{n_2} \\\\n_2 = \frac{n_1P_2}{P_1} \\\\n_2 = \frac{1.39 \times 7.8}{2.3} \\\\n_2 = 4.714 \ moles[/tex]
The mass of the oxygen in grams is calculated as;
m₂ = 4.714 x 32g
m₂ = 150.85 g
Therefore, the mass of oxygen the container must have is 150.85 g.
