Question

Graph between $\text{log}\frac{x}{m}$ and $\text{log P}$ is a straight line inclined at an angle $\theta ={45}^{\circ }$. When pressure is 0.5 atm and log K = 0.699, the amount of solute adsorbed per g of adsorbent will be:

• A. 0.397 g/g absorbent
• B. 1.5 g/g absorbent
• C. 2.5 g/g absorbent
• D. 0.25 g/g absorbent

Answer 1

The correct option is C

2.5 g/g absorbent

$\text{log}\frac{x}{m}=\text{log K}+\frac{1}{n}\text{log P}$

$\text{log}\frac{x}{m}=0.699+\left(\text{tan}45\right)\text{log}\left(0.5\right)$

$\text{log}\frac{x}{m}=0.3979$

$\frac{x}{m}=2.5$

$x=2.5\text{g}/\text{g}\text{adsorbent}$.