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

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(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

$200$

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(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 $gm$)

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