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

The molecules contain like 54 of water 

All the elements are arranged by melting points on the periodic table. in row they go from metals to gasses. in the row from sodium to argon, sodium is more reactive than the one on its right, magnesium.