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(B) According to the Freundlich adsorption isotherm: $\frac{x}{m} = kP^{1/n}$.Taking logarithm on both sides: $\log (x/m) = \log k + \frac{1}{n} \log

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$2$

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(B) According to the Freundlich adsorption isotherm: $\frac{x}{m} = kP^{1/n}$.Taking logarithm on both sides: $\log (x/m) = \log k + \frac{1}{n} \log P$.This equation represents a straight line $y = mx + c$,where the slope is $\frac{1}{n}$ and the intercept is $\log k$.Given slope $= 	an 45^{\circ} = 1$,so $\frac{1}{n} = 1$.Given intercept $= \log k = 0.3$. Since $\log 2 \approx 0.301$,we have $k = 2$.At $P = 1 \ atm$,the amount adsorbed $\frac{x}{m} = kP^{1/n} = 2 	imes (1)^1 = 2$.

In an adsorption experiment,a graph between $\log (x/m)$ vs $\log P$ was found to be linear with a slope of $45^{\circ}$. The intercept on the $\log (x/m)$ axis was found to be $0.3$. The amount of the gas adsorbed per gram of charcoal under a pressure of $1 \ atm$ is:

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