A sample of charcoal weighing 6.00 g was brought into contact with a gas contained in a vessel of 1.52 litre capacity at 27°C. The pressure of the gas was found to fall from 700 to 400 mm of Hg. The v

Question

A sample of charcoal weighing 6.00 g was brought into contact with a gas contained in a vessel of 1.52 litre capacity at 27°C. The pressure of the gas was found to fall from 700 to 400 mm of Hg. The volume of the gas (in ml), reduced to 0°C and 1 atm, that is adsorbed per gram of adsorbent under the experimental condition, is (Neglect the volume of the solid.)

Prashant Kotian · Accepted Answer

Correct answer: 0091. The drop in pressure due to adsorption is ΔP = 700 - 400 = 300 mm Hg = 300/760 atm. Using the ideal gas law to find the moles of gas adsorbed: Δn = (ΔP·V) / RT = [(300/760) atm × 1.52 L] / [0.0821 L·atm/(mol·K) × 300 K]. (300/760) × 1.52 = 0.6. Thus, Δn = 0.6 / 24.63 ≈ 0.02436 mol. At STP (0°C, 1 atm), the volume of 1 mole of gas is 22400 ml. Total volume of gas adsorbed at STP = 0.02436 mol × 22400 ml/mol = 545.6 ml. Volume adsorbed per gram of charcoal = 545.6 ml / 6.00 g = 90.93 ml/g ≈ 91 ml/g. Therefore, correct answer is 0091.

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