Question

A gas mixture is made by combining 6.16.1 g each of Ar, Ne, and an unknown diatomic gas. At STP, the mixture occupies a volume of 12.1212.12 L.

What is the molar mass of the unknown gas?

Identify the unknown gas.

Answer 1

Mass of Ar = 6.1 g

Moles of Ar = mass/molecular weight

= (6.1 g) / (40 g/mol)

= 0.1525 mol

Mass of Ne = 6.1 g

Moles of Ne = mass/molecular weight

= (6.1 g) / (20 g/mol)

= 0.305 mol

Mass of unknown gas = 6.1 g

Moles of unknown gas = mass/molecular weight

= 6.1/M

Total Moles = 0.1525 + 0.305 + (6.1/M) = 0.4575 + 6.1/M

At STP

Volume occupied = 22.4 L/mol

Total volume = (22.4 L/mol) x (0.4575 + 6.1/M) mol

12.12 L = (22.4 L/mol) x (0.4575 + 6.1/M) mol

0.541 = 0.4575 + 6.1/M

0.08357 = 6.1/M

M = 72.99 = 73 g/mol

Unknown gas is Cl2