CH4 is adsorbed on 1g charcoal at 0ºC following the Freundlich adsorption 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 a

Question

CH4 is adsorbed on 1g charcoal at 0ºC following the Freundlich adsorption 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____________(Nearest integer) x 10–2 [Use log102 = 0.3010 log103 = 0.4771] 0ºC ij CH4 dk 1 g pkjdksy ij vf/k'kks"k.k] Qzk;UMfyd vf/k'kks"k.k lerkih dk vuqlj.k djrk gSA mm of Hg, ij 10.0 mL CH4 vf/k'kksf"kr gks tkrh gS] tcfd 200 mm Hg ij 15.0 mL vf/k'kksf"kr gksrh gSA 300 mm of Hg ij CH4 ds vf/k'kksf"kr vk;ru dk eku 10x mL gSA x dk eku____________ x 10–2 ¼fudre iw.kkZd esa½ [mi;ksx dhft, log102 = 0.3010 log103 = 0.4771]

Prashant Kotian · Accepted Answer

Correct answer: .. n Kp m x  n ) ( K 10  ………….(1) n ) ( K 15  ………….(2) n ) ( K V  ………….(3) Divide (2) to (1) = n 2 15  log n 2 log        . 3010 . 3010 . 4771 . n    Divide (3) to (1) = n 3 V  log n 10 V log        . 4771 . 585 . 10 V log          . 10 V  V = 10 × 100.2791 = 101.2791=10x x = 1.2791 = 127.91 × 10-2  128 × 10-2

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