A rigid container has 44.5 grams of oxygen gas at room temperature and a pressure of 2.3 atm. How many grams of oxygen should the container have for the pressure to be 7.8 atm?

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

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.

the answer is 0.42 is that is an option. this question provides no info