Some CO₂ gas was kept in a sealed container at a pressure of 1 atm and at 273 K. This entire amount of CO₂ gas was later passed through an aqueous solution of Ca(OH)₂. The excess unreacted Ca(OH)₂ was later neutralized with 0.1 M of 40 mL HCl. If the volume of the sealed container of CO₂ was x, then x is _______ cm³ (nearest integer). [Given : The entire amount of CO₂(g) reacted with exactly half of the initial amount of Ca(OH)₂ present in the aqueous solution.]

Question

Dr. Abubacker · Accepted Answer

[Solution by Dr. Abubacker] 0.1 M × 40 mL = 4 mmol HCl neutralises 2 mmol of excess Ca(OH)₂. CO₂ consumed exactly half of the initial Ca(OH)₂, so the amount consumed equals the excess: 2 mmol. Hence n(CO₂) = 2 mmol and, at 1 atm and 273 K, V = 2 × 22.4 = 44.8 cm³ ≈ 45. Answer: 45.

Explanation

Reviewed by Dr. Abubacker

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