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(B) The Freundlich adsorption isotherm is given by $\frac{x}{m} = KP^{1/n}$. Since the mass of adsorbent $(m)$ is constant $(1 \ g)$,we have $V \propt

Vedclass · Accepted Answer

$128$

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(B) The Freundlich adsorption isotherm is given by $\frac{x}{m} = KP^{1/n}$. Since the mass of adsorbent $(m)$ is constant $(1 \ g)$,we have $V \propto \frac{x}{m}$,so $V = KP^{1/n}$.For the given data:$10 = K(100)^{1/n} \quad (1)$$15 = K(200)^{1/n} \quad (2)$$V = K(300)^{1/n} \quad (3)$Dividing $(2)$ by $(1)$:$\frac{15}{10} = \left(\frac{200}{100}ight)^{1/n} \Rightarrow 1.5 = 2^{1/n}$.Taking $\log_{10}$ on both sides:$\log_{10} 1.5 = \frac{1}{n} \log_{10} 2$$\log_{10} \left(\frac{3}{2}ight) = \frac{1}{n} (0.3010)$$0.4771 - 0.3010 = \frac{1}{n} (0.3010)$$0.1761 = \frac{1}{n} (0.3010) \Rightarrow \frac{1}{n} = \frac{0.1761}{0.3010} \approx 0.585$.Dividing $(3)$ by $(1)$:$\frac{V}{10} = \left(\frac{300}{100}ight)^{1/n} = 3^{1/n}$.Taking $\log_{10}$ on both sides:$\log_{10} \left(\frac{V}{10}ight) = \frac{1}{n} \log_{10} 3$$\log_{10} \left(\frac{V}{10}ight) = 0.585 	imes 0.4771 = 0.2791$.$\frac{V}{10} = 10^{0.2791} \Rightarrow V = 10 	imes 10^{0.2791} = 10^{1.2791}$.Given $V = 10^x$,so $x = 1.2791$.Rounding to the nearest integer for $x 	imes 10^{-2}$,we get $128 	imes 10^{-2}$.

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