Question

A plot of $\text{log}\frac{x}{m}$ vs. $\text{log p}$ for adsorption of a gas on to a solid gives us a straight line with slope equal to:

• A. $n$
• B. $\frac{1}{n}$
• C. $\text{log K}$
• D. $\text{log p}$

Answer 1

The correct option is B $\frac{1}{n}$

According to Freundlich:
$\frac{x}{m}=K{p}^{\frac{1}{n}}$
or, $\text{log}\left(\frac{x}{m}\right)=\text{log K}+\frac{1}{n}\text{log}p$
Where,
${}^{\prime }{x}^{\prime }$ is the mass of adsorbate,
${}^{\prime }{m}^{\prime }$ is the mass of adsorbent,
${}^{\prime }{K}^{\prime }$ and ${}^{\prime }{n}^{\prime }$ are constants which depend on the nature of the adsorbent and the gas at a particular temperature. ${}^{\prime }{p}^{\prime }$ is the pressure.