A0.334 g sample of an unknown halogen occupies 109 ml at 398 k and 1.41 atm. what is the identity of the halogen?

We can use the ideal gas law equation for the above reaction to find the number of moles present 
PV = nRT 
P - pressure - 1.41 atm x 101325 Pa/atm = 142 868 Pa
V - 109 x 10⁻⁶ m³
R - 8.314 Jmol⁻¹K⁻¹
T - 398 K
substituting the values in the equation 
142 868 Pa x 109 x 10⁻⁶ m³ = n x 8.314 Jmol⁻¹K⁻¹ x 398 K
n = 4.70 x 10⁻³ mol
number of moles = mass present / molar mass
molar mass = mass / number of moles  
                   = 0.334 g/ 4.70 x 10⁻³ mol = 71.06 g/mol
halogens exist as diatomic molecules 
Therefore atomic mass - 71.06 / 2 = 35.5 
halogen with 35.5 g/mol is Cl
unknown halogen is Cl