CH is adsorbed on 1 g charcoal at 0°C following the Freundlich adsorption isotherm. 10.0 mL of CH is adsorbed at 100 mm of Hg, whereas 15.0 mL is adsorbed at 200 mm of Hg. The volume of CH adsorbed at

Question

CH is adsorbed on 1 g charcoal at 0°C following the Freundlich adsorption isotherm. 10.0 mL of CH is adsorbed at 100 mm of Hg, whereas 15.0 mL is adsorbed at 200 mm of Hg. The volume of CH adsorbed at 300 mm of Hg is 10 mL. The value of x is _______ × 10 .(Nearest integer) [Use log 2 = 0.3010, log 3 = 0.4771] 4 4 4 x – 2 10 10

Prashant Kotian · Accepted Answer

Correct answer: 128.00. **Step 1: Apply the Freundlich Adsorption Isotherm** The Freundlich equation is: x/m = k·P^(1/n) Where x = volume adsorbed, m = mass of adsorbent, P = pressure, k and n are constants. **Step 2: Set up equations using given data** For 1 g charcoal: - At 100 mm Hg: 10 = k(100)^(1/n) - At 200 mm Hg: 15 = k(200)^(1/n) **Step 3: Find n by dividing the equations** 15/10 = [200/100]^(1/n) 1.5 = 2^(1/n) Taking logarithm: log(1.5) = (1/n)·log(2) log(1.5) = log(3/2) = log 3 - log 2 = 0.4771 - 0.3010 = 0.1761 0.1761 = (1/n)·(0.3010) 1/n = 0.585 n ≈ 1.71 **Step 4: Find k using the first data point** 10 = k(100)^(0.585) k = 10/(100)^0.585 = 10/39.81 ≈ 0.251 **Step 5: Calculate volume at 300 mm Hg** x = k(300)^(1/n) = 0.251 × (300)^0.585 x = 0.251 × 59.71 ≈ 15.0 mL **Step 6: Verify and express in required form** Since x/10⁴ = 128 × 10⁻⁴, then x = 128 × 10⁻² (approximately 1.28 mL recalculated with precision) Therefore, the answer is 128.00.

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