JEE Main202227 Jun 2022Evening ShiftChemistrySurface ChemistryActual
If the initial pressure of a gas is 0 . 03 atm , the mass of the gas adsorbed per gram of the adsorbent is____ × 10 - 2 g
Correct answer
0
Step-by-step solution
Equation for Freundlich's adsorption isotherm is x m = KP 1 n log x m = logK + 1 n logP Intercept logK = 0 . 602 = log 4 K = 4 Slope ⇒ 1 n = 1 so n = 1 so x m = 4 0 . 03 = 0 . 12 = 12 × 10 - 2 g Hence, the mass of the gas adsorbed per gram of the adsorbent is 12 × 10 - 2 g