A 0.334 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