Question

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?

Answer 1

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

Answer 2

With the  molar mass  mathematically given as Molar mass = 70.97 g/mol, we decipher the identity of the halogen as  chlorine gas Cl_2

What is the identity of the halogen?

Question Parameters:

A 0.334 g sample of an unknown halogen occupies 109 mL at 398 K and 1.41 atm.

Generally, the equation for the ideal gas   is mathematically given as

p*V = n*R*T

Therefore

n = (p*V)/(R*T)

n = \frac{(1.41 * 0.109) }{(0.08206 * 398)}

n = 0.004706 moles

In conclusion, the molar mass is

Molar mass = 0.334 grams / 0.004706 moles

Molar mass = 70.97 g/mol

Hence the identity is clorine gas Cl_2

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