When a graph is plotted between logxm and log p- it is a straight line with an angle 45- and intercept 0-3010 on y-axis- If initial pressure is 0-3 atm- what will be the amount of gas adsorbed per g of adsorbent-

According to nbsp;Freundlich adsorption isotherm, we have logxm=log k+1nlog p ∴ Slope is 1n=tan 45^∘ ⇒ n=1 log k=0.3010 ⇒ k=2 We know that nbsp; xm= k. p^1/ ...

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Standard XII

Chemistry

Types of Catalysis

When a graph ...

Question

When a graph is plotted between $log \frac{x}{m}$ and $log p$ , it is a straight line with an angle $45^{\circ}$ and intercept $0.3010$ on $y -$ axis. If initial pressure is $0.3 atm$ . what will be the amount of gas adsorbed per $g$ of adsorbent:-

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Solution

The correct option is B $0.6$
According to Freundlich adsorption isotherm, we have
$log \frac{x}{m} = log k + \frac{1}{n} log p$
$∴$ Slope is $\frac{1}{n} = tan 45^{\circ} \Rightarrow n = 1$
$log k = 0.3010$
$\Rightarrow k = 2$
We know that
$\frac{x}{m} = k . p^{1 / n}$
$\Rightarrow \frac{x}{m} = 2 \times (0.3)^{1} = 0.6$
As we need to calculate per $g$ of adsorbent, then
$x = 0.6$

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When a graph is plotted between logxm and logp, it is a straight line with an angle 45∘ and intercept 0.3010 on y−axis. If initial pressure is 0.3 atm. what will be the amount of gas adsorbed per g of adsorbent:-

The correct option is B 0.6According to Freundlich adsorption isotherm, we have logxm=logk+1nlogp ∴ Slope is 1n=tan45∘⇒n=1 logk=0.3010 ⇒k=2 We know that xm=k.p1/n ⇒xm=2×(0.3)1=0.6 As we need to calculate per g of adsorbent, then x=0.6

When a graph is plotted between $log \frac{x}{m}$ and $log p$ , it is a straight line with an angle $45^{\circ}$ and intercept $0.3010$ on $y -$ axis. If initial pressure is $0.3 atm$ . what will be the amount of gas adsorbed per $g$ of adsorbent:-