(B) According to the Freundlich adsorption isotherm,$\frac{x}{m} = k \cdot p^{1/n}$.Taking $\log$ on both sides,we get $\log (\frac{x}{m}) = \log k +

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(B) According to the Freundlich adsorption isotherm,$\frac{x}{m} = k \cdot p^{1/n}$.Taking $\log$ on both sides,we get $\log (\frac{x}{m}) = \log k + \frac{1}{n} \log p$.Comparing this with the equation of a straight line $y = mx + c$,the slope is $\frac{1}{n} = \tan 45^\circ = 1$,so $n = 1$.The intercept is $\log k = 0.3010$. Since $\log 2 = 0.3010$,we have $k = 2$.Substituting these values at $p = 0.5 \ atm$,we get $\frac{x}{m} = 2 \times (0.5)^1 = 1.0 \ g$.

Class 12 Chemistry Surface Chemistry Adsorption and Adsorption isotherm

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In an adsorption experiment,a graph between $\log (x/m)$ versus $\log P$ was found to be linear with a slope of $45^\circ$. The intercept on $\log (x/m)$ axis was found to be $0.3010$. The amount of gas adsorbed per gram of charcoal under a pressure of $0.5 \ atm$ is (in $g$)

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A
$0.5$
B
$1.0$
C
$1.5$
D
$0.75$

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Surface Chemistry

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Adsorption and Adsorption isotherm

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In an adsorption experiment,a graph between $\log (x/m)$ versus $\log P$ was found to be linear with a slope of $45^\circ$. The intercept on $\log (x/m)$ axis was found to be $0.3010$. The amount of gas adsorbed per gram of charcoal under a pressure of $0.5 \ atm$ is (in $g$)

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$3.12 \, g$ of oxygen is adsorbed on $1.2 \, g$ of platinum metal. The volume of oxygen adsorbed per gram of the adsorbent at $1 \, atm$ and $300 \, K$ in $L$ is ..... .$[R = 0.0821 \, L \, atm \, K^{-1} \, mol^{-1}]$

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$3.12 \, g$ of oxygen is adsorbed on $1.2 \, g$ of platinum metal. The volume of oxygen adsorbed per gram of the adsorbent at $1 \, atm$ and $300 \, K$ in $L$ is ..... .
$[R = 0.0821 \, L \, atm \, K^{-1} \, mol^{-1}]$