The graph of log frac{x}{m} vs log p for an a | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The Freundlich adsorption isotherm is given by the equation: $\log \frac{x}{m} = \log k + \frac{1}{n} \log P$. Comparing this with the equation of

Vedclass · Accepted Answer

$16$

Vedclass · Answer

(D) The Freundlich adsorption isotherm is given by the equation: $\log \frac{x}{m} = \log k + \frac{1}{n} \log P$. Comparing this with the equation of a straight line $y = mx + c$,we have the slope $\frac{1}{n} = 	an 45^{\circ} = 1$ and the intercept $c = \log k = 0.6020$. Since $\log 2 = 0.3010$,then $2 	imes \log 2 = 0.6020$,which implies $\log 2^2 = \log 4 = 0.6020$. Thus,$k = 4$. Substituting these values into the isotherm equation: $\frac{x}{m} = k \cdot P^{1/n} = 4 	imes (0.4)^1 = 1.6$. Expressing $1.6$ in the form $......... 	imes 10^{-1}$,we get $1.6 = 16 	imes 10^{-1}$. Therefore,the nearest integer is $16$.

Similar Questions

Give a reason why a finely divided substance is more effective as an adsorbent.

The Langmuir adsorption isotherm is based on which of the following assumptions?

Which one of the following graphs represents the Freundlich adsorption isotherm?

Which one of the following characteristics is associated with adsorption?

The plot that best represents the relationship between the extent of adsorption $(x / m)$ and pressure $(p)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module