Question

At $T(K)$ adsorption of acetic acid on 1g of charcoal gave a Freundlich adsorption isotherm with slope of 0.5 and intercept of 1. What is the value of x, when the concentration of acetic acid is 0.1 mol $L^{-1}$? (Given: $\text{antilog}(0.5) [cite_start]= 3.162$; $\text{antilog}(0.301) = 2.0$)

• A. 0.3162
• B. 3.162
• C. 0.5
• D. 0.2

Hint:

Don't let log forms overcomplicate things. Work with the base equation $\frac{x}{m} = k \cdot C^{1/n}$ once you extract $k$ and $\frac{1}{n}$.

Answer 1

The Correct Option is B

Step 1: Concept
The Freundlich adsorption isotherm describes how a gas or solute adheres to a solid surface: $\frac{x}{m} = k \cdot C^{1/n}$, where $x$ is the mass of adsorbate, $m$ is the mass of adsorbent, and $C$ is the concentration.

Step 2: Meaning
Taking the logarithm of the Freundlich equation transforms it into a linear equation ($y = mx + c$): $$\log\left(\frac{x}{m}\right) = \log k + \frac{1}{n}\log C$$ Here, the slope is $\frac{1}{n} = 0.5$ and the y-intercept is $\log k = 1$.

Step 3: Analysis
Given details: * Mass of adsorbent ($m$) = $1\text{ g}$ * Intercept $\log k = 1 \implies k = \text{10}^{1} = 10$ * Slope $\frac{1}{n} = 0.5$ * Concentration ($C$) = $0.1\text{ mol L}^{-1}$ Substitute these parameters back into the foundational Freundlich form: $$\frac{x}{1} = 10 \cdot (0.1)^{0.5}$$ $$x = 10 \cdot \sqrt{0.1} = 10 \cdot \sqrt{\frac{1}{10}} = \frac{10}{\sqrt{10}} = \sqrt{10}$$ Using the log tables provided: $\text{antilog}(0.5) = 3.162$, which represents $\text{10}^{0.5} = \sqrt{10} = 3.162$. $$x = 3.162\text{ g}$$

Step 4: Conclusion
This calculation yields a value of 3.162.

Final Answer: (B)