Adsorption of phenol from its aqueous solutio | vedclass

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Question

$\frac{ x }{ m }= K 	imes C ^{1 / n }$
$\log \left(\frac{x}{m}ight)=\log K+\frac{1}{n} \log C$
$\log 4=\log K+\frac{1}{n} \log 10$
$0.6=\log K+\

Vedclass · Accepted Answer

$15.62$ or $16.00$

Vedclass · Answer

$\frac{ x }{ m }= K 	imes C ^{1 / n }$
$\log \left(\frac{x}{m}ight)=\log K+\frac{1}{n} \log C$
$\log 4=\log K+\frac{1}{n} \log 10$
$0.6=\log K+\frac{1}{n}...(1)$
$\log 10=\log K+\frac{1}{n} \log 16$
$1=\log K+\frac{1}{n} 	imes 1.2...(2)$
Equation $(2) -$ equation $(1)$
$0.4=\frac{1}{n} 	imes(0.2) \Rightarrow n=0.5$
and $\log K =-1.4$
$\log \frac{x}{m}=\log K+\frac{1}{n} 	imes \log C$
$=-1.4+2 	imes \log 20$
$=-1.4+2.6=1.2$
$\frac{x}{m}=10^{+1.2}=16$
$\left(\log 2=0.3,4 \log 2=1.2,16=10^{+1.2}ight)$
or
$\frac{x}{m}=K 	imes C^{1 / n}$
$4=K(10)^{1 / n}....(1)$
$10=K(16)^{1 / n}.....(2)$
$X=K(20)^{1 / n}.......(3)$
On solving equation $(1)$ and $(2)$
$\frac{1}{n}=2$
On solving equation $(1)$ and $(3)$
$\frac{4}{X}=\left(\frac{10}{20}ight)^2$
$X=16$
On solving equation $(2)$ and $(3)$
$\frac{10}{X}=\left(\frac{16}{20}ight)^2$
$X=15.625$

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