At s.t.p, a mole of chlorine occupies 22.4dm3. If the temperature is changed to 35°C and the pressure to 750mHg the of the new volume will be [Cl = 35.5]

Analyze the question: One mole of chlorine gas occupies 22.4 dm³ at s.t.p. (760 mmHg, 273 K). The conditions change to 35°C and 750 mmHg, and we need the new volume. This is a combined gas law problem.

Concept: The combined gas law says

P₁V₁/T₁ = P₂V₂/T₂

Temperature must be in Kelvin.

Solution:

• P₁ = 760 mmHg, V₁ = 22.4 dm³, T₁ = 273 K
• P₂ = 750 mmHg, V₂ = ?, T₂ = 35 + 273 = 308 K

Make V₂ the subject:

V₂ = (P₁ × V₁ × T₂) / (P₂ × T₁)

V₂ = (760 × 22.4 × 308) / (750 × 273) dm³

Checking the options:

• The molar mass of Cl (35.5) is not needed, because the volume per mole is already given. This rules out A, C, D and E, which all include 35.5.
• Option B matches our result exactly.

Sense check: the temperature rises (the volume increases) and the pressure drops slightly (the volume increases), so V₂ should be greater than 22.4 dm³. The factor (760 × 308)/(750 × 273) is greater than 1, so this is consistent.

Answer: B

Tip: Always convert °C to K first (add 273). Also watch for distractors like the atomic mass. If the question already gives the volume of one mole, the molar mass is not needed. In the combined gas law, put the "new" temperature on top and the "new" pressure on the bottom.