**If the initial pressure of a gas is 0.03 atm, the mass of the gas adsorbed per gram of the adsorbent is____________* 10^-2g.**

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Payal Gupta | Contributor-Level 10

7 months ago

Using, Freundlich adsorption isotherm ;

$\frac{x}{m} = k . p^{1 / n}$ ………………. (i)

$l o g \frac{x}{m} = \frac{1}{n} l o g p + l o g k$

Comparing with y = mx + C

Slope = $\frac{1}{n}$ = 1

Intercept, log k = 0.602

log k = log 4

k = 4

from equation (i)

$\frac{x}{m} = 4 * {(0.03)}^{1}$

= 0.12

= 12 * 10^-²

So, 12 * 10^-² g of gas is adsorbed per gram of adsorbent,

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If the initial pressure of a gas is 0.03 atm, the mass of the gas adsorbed per gram of the adsorbent is____________* 10-2g.

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