The graph between log (frac{x}{m}) and log P | vedclass

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

Question

(C) According to the Freundlich adsorption isotherm,$\log (\frac{x}{m}) = \log K + \frac{1}{n} \log P$.From the given graph,the slope is $	an(45^{\ci

Vedclass · Accepted Answer

$200$

Vedclass · Answer

(C) According to the Freundlich adsorption isotherm,$\log (\frac{x}{m}) = \log K + \frac{1}{n} \log P$.From the given graph,the slope is $	an(45^{\circ}) = 1$,so $\frac{1}{n} = 1$.The intercept on the y-axis is $\log K = 0.3$. Since $\log 2 \approx 0.301$,we have $K = 2$.Substituting these values into the Freundlich equation: $\frac{x}{m} = K \cdot P^{1/n} = 2 \cdot P^1$.Given $m = 50 \ gm$ and $P = 2 \ atm$,we get $\frac{x}{50} = 2 \cdot 2 = 4$.Therefore,$x = 50 \cdot 4 = 200 \ gm$.

The graph between $\log (\frac{x}{m})$ and $\log P$ is provided for the adsorption of $NH_3$ gas on a metal surface. Calculate the weight of $NH_3$ gas adsorbed by $50 \ gm$ of the metal surface at $2 \ atm$ pressure. (in $gm$)

Similar Questions

Which of the following statements is incorrect w.r.t. $ \text{Physisorption} $?

Why is chemisorption referred to as activated adsorption?

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

Which of the following properties belongs to physical adsorption?

The statement not applicable to chemisorption is

Vedclass Test Series

Exam Paper Generator

Online Exam Module