JEE Main · 2020 · 06 Sep 2020 · Shift 2 · Surface Chemistry

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For Freudlich adsorption isotherm, a plot of log(x/m) (y-axis) and log p (x-axis) gives a straight line. The intercept and slope for the line is 0.4771 and 2, respectively. The mass of gas adsorbed per gram of adsorbent if the initial pressure is 0.04 atm, is _____ × 10⁻⁴ g. (log 3 = 0.4771)

Answer: 48

Solution

The Freundlich isotherm

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

becomes

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

on taking logs, so the intercept gives

lo g k = 0.4771

, i.e.

k = 3

, and the slope gives

\frac{1}{n} = 2

. At

P = 0.04 atm

\frac{x}{m} = 3 \times (0.04)^{2} = 3 \times 1.6 \times 1 0^{- 3} = 4.8 \times 1 0^{- 3} g

, which is

48 \times 1 0^{- 4} g

JEE Main · 2020 · 06 Sep 2020 · Shift 2 · Integer answer

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