JEE Main · 2022 · 27 Jun 2022 · Shift 2 · Surface Chemistry

If the initial pressure of a gas is 0.03 atm, the mass of the gas adsorbed per gram of the adsorbent is ______

\times 1 0^{- 2} g

Answer: 12

Solution

The Freundlich adsorption isotherm is:

\frac{x}{m} = k P^{1/ n}

Taking logarithm:

lo g \frac{x}{m} = lo g k + \frac{1}{n} lo g P

From the graph provided:

Substituting

P = 0.03

atm:

\frac{x}{m} = 4 \times (0.03)^{1} = 0.12 = 12 \times 1 0^{- 2}

JEE Main · 2022 · 27 Jun 2022 · Shift 2 · Integer answer