Suppose 0.508 g of a gas occupies a volume of 0.175 L at a temperature of 25.0◦C is held at a pressure of 1.0 atm. What gas is this?

PV = nRT
(1.0 atm) × (0.175 L) = n × [0.0821 atm L / (mol K)] × [(273.2 + 25.0) K]
n = 1.0 × 0.175 / (0.0821 × 298.2) mol
n = 0.00715 mol

Molar mass of the gas = (0.508 g) / (0.00715 mol) = 71.0 g/mol
The molar mass of Cl₂ (chorine) = 35.5 × 2 g/mol = 71.0 g/mol
Hence, the gas is chlorine.

The answer: B) Chlorine