CH₄ is absorbed on 1g charcoal at 0°C following the Freundlich adsorpiton isotherm 10.0 mL of CH₄ is adsorbed at 100 mm of Hg, whereas 15.0 mL is adsorbed at 200 mm of Hg. The volume of CH₄ adsorbed at 300 mm of Hg is 10ˣ mL. The value of x is _________ × 10⁻².
(Nearest integer) [Use log₁₀2 = 0.3010, log₁₀3 = 0.4771]

Solution

Chemistry ma’am · 4 steps

Take the ratio of two Freundlich readings so k cancels and you get 1/n, then reuse it at 300 mm.

1. Freundlich isotherm. For a fixed 1 g of charcoal, the volume adsorbed follows V = k p^(1/n). Dividing two readings removes k, so V₂/V₁ = (p₂/p₁)^(1/n).
2. Find 1/n. From 100 mm to 200 mm, 15/10 = 2^(1/n). So 1/n = (log 1.5)/(log 2) = (0.4771 - 0.3010)/0.3010 = 0.1761/0.3010 = 0.5850.

   log V₃ = log 10 + 1/n log 300/100 = 1 + 0.5850 × 0.4771

3. Volume at 300 mm. 0.5850 × 0.4771 = 0.2791, so log V₃ = 1.2791. Since V₃ = 10ˣ mL, x = 1.2791 = 127.9 × 10⁻².
4. So. V₃ ≈ 19 mL, and the value asked for is 127.9, which rounds to 128.

The common slip: 28. Taking the log of the ratio V₃/V₁ instead of V₃ itself drops the log 10 = 1 and gives 28, not 128.

The idea to keep. With two Freundlich readings, divide them so k cancels and solve for 1/n from the log ratio. Then compare any new pressure with a known reading the same way.

Answer from the NTA final answer key. Solution by PrepWiser.