Adsorption of phenol from its aqueous solutio | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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 $(3)$
$\frac{10}{X}=\left(\frac{16}{20}ight)^2$
$X=15.625$

Similar Questions

Which of the following orders is incorrect?

$A$ spring of spring constant $5 \times 10^3 \, N/m$ is stretched initially by $5 \, cm$ from the unstretched position. Then the work required to stretch it further by another $5 \, cm$ is .............. $N-m$.

In a series $LCR$ circuit,what will be the nature of the circuit for frequencies higher than the resonance frequency?

What is the chemical formula of Glauber's salt?

Breeding crops with higher levels of vitamins,minerals,and proteins is known as ..........

Vedclass Test Series

Exam Paper Generator

Online Exam Module