Question

Adsorption of a gas on a solid obeys the Freundlich adsorption isotherm. In the graph drawn between \(\log (x/m)\) (on the y-axis) and \(\log p\) (on the x-axis), the slope and intercept are found to be 1 and 0.3, respectively. If the initial pressure of the gas is 0.02 atm, the mass of the gas adsorbed per gram of solid is (\({antilog } 0.3 = 2\)). 
 

• A. \(2 \times 10^{-2}\) g
• B. \(3 \times 10^{-2}\) g
• C. \(4 \times 10^{-2}\) g
• D. \(6 \times 10^{-2}\) g

Hint:

In adsorption isotherm calculations, always ensure to correctly interpret the slope and intercept from the logarithmic plot and use appropriate antilogarithm conversions.

Answer 1

The Correct Option is C

The Freundlich adsorption isotherm is given by: \[ \log \left(\frac{x}{m}\right) = \log k + n \log p \] where: 
- \( n = 1 \) (slope), 
- \( \log k = 0.3 \) (intercept), 
- \( p = 0.02 \) atm. Substituting the values: \[ \log \left(\frac{x}{m}\right) = 0.3 + 1 \times \log(0.02) \] Since \(\log (0.02) = -1.7\): \[ \log \left(\frac{x}{m}\right) = 0.3 - 1.7 = -1.4 \] Taking the antilog: \[ \frac{x}{m} = {antilog} (-1.4) = \frac{2}{10^{1.4}} \] Approximating \(10^{1.4} \approx 25\): \[ \frac{x}{m} = \frac{2}{25} = 0.04 \] Thus, the mass of gas adsorbed per gram of solid is \(4 \times 10^{-2}\) g.