Question

Plot of $\log(\frac{x}{m})$ against log P is a straight line with a slope of $\frac{1}{2}$. When the pressure is 0.5 atm and Freundlich parameter, k is 10. The amount of solute adsorbed per gram of adsorbent will be.

• A. 1 g
• B. 2 g
• C. 3 g
• D. 7 g

Answer 1

Answer: (D)

7 g

Answer 2

The Freundlich adsorption isotherm is given by:

$\frac{x}{m} = k \, p^{1/n}$

Taking the logarithm on both sides:

$\log \frac{x}{m} = \log k + \frac{1}{n} \log p$

Here, the slope of the plot ($\log \frac{x}{m}$ vs $\log p$) is $\frac{1}{n}$. Given that the slope is $\frac{1}{2}$, we have:

$\frac{1}{n} = \frac{1}{2} \quad \Rightarrow \quad n = 2$

Now, using the equation with $k = 10$ and $p = 0.5 \, \text{atm}$:

$\frac{x}{m} = 10 \, p^{1/2} = 10 \sqrt{0.5}$

Since $\sqrt{0.5} = \sqrt{\frac{1}{2}} \approx 0.7071$, we get:

$\frac{x}{m} \approx 10 \times 0.7071 = 7.071 \, \text{g}$

Thus, the amount of solute adsorbed per gram of adsorbent is approximately 7 g.