JEE Mains Chemistry Past PapershardNUMERICAL

The graph of log m x vs log p for an adsorption process is a straight line inclined at an angle of 4JEE Mains Chemistry Past Papers Chemistry Question

Question

The graph of log m x vs log p for an adsorption process is a straight line inclined at an angle of 45º with intercept equal to 0.6020. The mass of gas adsorbed per unit mass of adsorbent at the pressure of 0.4 atm is ______ × 10–1 (Nearest integer). Given: log 2 = 0.3010

Answer: .

💡 Solution & Explanation

n kP m x  log(x/m) = logk + n 1 logP log (x/m) vs logP graph will be straight ling with slope = n 1 = tan45º = 1 y = intercept = log k1 = 0.6020 = log 4  k = 4 log m x = log4 + log 0.4 log m x = log 1.6 m x = 1.6  16 × 10–1 | JEE(Main) 2023 | DATE : 30-01-2023 (SHIFT-2) | PAPER-1 | CHEMISTRY PAGE # 9

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