JEE PYQ: Surface Chemistry Question 20
Question 20 - 2020 (06 Sep Shift 2)
Type: Numerical
For Freundlich 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 ____ $\times 10^{-4}$ g. ($\log 3 = 0.4771$)
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Answer: 48
Solution
Freundlich adsorption isotherm: $\frac{x}{m} = k \cdot p^{1/n}$
$\log \frac{x}{m} = \log k + \frac{1}{n} \log p$
Slope $\frac{1}{n} = 2$, Intercept $= \log k = 0.4771$, so $k = \text{Antilog}(0.4771) = 3$
$\frac{x}{m} = k(p)^{1/n} = 3 \times (0.04)^2 = 3 \times 0.0016 = 48 \times 10^{-4}$
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