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? R = 0.0821 퐿.푎푡푚푚표푙.퐾

The halogen is the Cl₂ (chlorine)

Explanation:

Let's solve this by the Ideal Gases Law → P . V = n . R . T

We need the molar mass to identify the halogen

1.41 atm . 0.109L = n. 0.082 . 398K

(1.41 atm . 0.109L) / 0.082 . 398K = n

4.71×10⁻³ = mol

Let's find out the molar mass → 0.334 g / 4.71×10⁻³mol = 70.9 g/mol

The halogen is the Cl₂

The halogen is Cl2

Explanation:

Step 1: Data given

Mass of the halogen = 0.334 grams

Volume of the halogen = 109 mL = 0.109 L

Temperature = 398 K

Pressure = 1.41 atm

Step 2: Calculate moles of halogen

p*V = n*R*T

⇒ p = the pressure = 1.41 atm

⇒ V = the volume = 0.109 L

⇒n = the number of moles = TO BE DETERMINED

⇒ R = the gas constant = 0.08206 L*atm/mol*K

⇒T = the temperature = 398 K

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

n = (1.41 * 0.109) / (0.08206 * 398)

n = 0.004706 moles

Step 3: Calculate molar mass

Molar mass = mass / moles

Molar mass = 0.334 grams / 0.004706 moles

Molar mass = 70.97 g/mol

The halogen is Cl2

the answer is a and have a great day

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