calculate the volume of 45.0 grams of carbon dioxide gas at 280 K and 1.2 atm?

Molar mass of CO₂ = (12.0 + 16.0×2) g/mol = 44.0 g/mol

For the sample of CO₂ :
No. of moles, n = 45.0/44.0 mol
Temperature, T = 280 K
Pressure, P = 1.2 atm
Gas constant, R = 0.08206 L atm / (mol K)

For a fixed amount of gas : PV = nRT
Then, V = nRT/P

Volume of CO₂, V = (45.0/44.0) × 0.08206 × 280 / 1.2 = 19.6 L

PV = nRT

P = 1.2 atm
T = 280 K
n = 45.0 g / the molecular weight of CO2 (this puts the answer in moles)
R = 0.08206 L-atm/deg-mol

Solve for V

V = nRT/P

im not sure but my guess is 43.4