Question 60

A Freundlich adsorption isotherm graph is shown.

If the initial pressure of a gas is $$0.03$$ atm, the mass of the gas adsorbed per gram of the adsorbent is ______ $$\times 10^{-2}$$ g

Correct Answer: 12

Solution

1. Identify Graph Information

The straight-line equation for the Freundlich isotherm is given by:

$$\log(x/m) = \log k + (1/n) \log p$$

From the provided line graph data, we can match the parameters directly:

Slope (1/n): 1
Intercept (log k): 0.602
Initial Pressure (p): 0.03 atm

2. Calculate Adsorption Constant (k)

Using the intercept value to solve for k:

$$log k = 0.602$$

Since $$\log(4)\approx0.602$$

(because $$log 2 ≈ 0.301$$):

$$k = 10^{0.602} = 4$$

3. Solve for Mass Adsorbed per Gram (x/m)

Substitute the constants back into the base equation

$$x/m = k · p^{1}$$

$$x/m = 4 · (0.03)^1$$

$$x/m = 0.12 g$$

4. Format as Scientific Notation

$$0.12 g = 12 × 10^{−2}$$ g

Correct Integer Answer: 12

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A Freundlich adsorption isotherm graph is shown. If the initial pressure of a gas is $$0.03$$ atm, the mass of the gas adsorbed per gram of the adsorbent is ______ $$\times 10^{-2}$$ g