How many grams of chlorine gas (cl2) are in a 17.8-liter sample at 1.1 atm and 29 c?

We can use the ideal gas law equation to find the number of Cl moles in the container 
PV = nRT
where
P - pressure - 1.1 atm x 101 325 Pa/atm = 111 458 Pa
V - 17.8 x 10⁻³ m³
n - number of moles
R - universal gas constant 8.314 Jmol⁻¹K⁻¹
T - temperature in Kelvin - 273 + 29 °C = 302 K 
substituting these values in the equation 
111 458 Pa x 17.8 x 10⁻³ m³ = n x 8.314 Jmol⁻¹K⁻¹ x 302 K
n = 0.790 mol
molar mass of Cl₂ = 71 g/mol
therefore mass of Cl₂ = 0.790 mol x 71 g/mol = 56.1 g
mass of Cl₂ present - 56.1 g

The answer is 323 g.