1

### JEE Main 2018 (Online) 15th April Evening Slot

If x gram of gas is adsorbed by m gram of adsorbent at pressure P, the plot of log ${x \over m}$ versus log P is linear. The slope of the plot is :
(n and k are constants and n > 1)
A
2 k
B
log k
C
n
D
${1 \over n}$

## Explanation

According to Freundlich adsorption isotherm, in the median range of pressure

${x \over m} \propto {P^{{1 \over n}}}$

$\Rightarrow$  ${x \over m}$ = kP$^{{1 \over n}}$

taking log both sides, we get,

log${x \over m}$ = logk + ${1 \over n}$ logP

Here in graph between log${x \over m}$ and logP, slope is ${1 \over n}$ and intercepts = log k.
2

### JEE Main 2018 (Online) 16th April Morning Slot

Which one of the following is not a property of physical adsorption ?
A
Higher the pressure, mre the adsorption
B
Lower the temperature, more the adsorption
C
Greater the surface area, more the adsorption
D

## Explanation

Physical absorption increases with increase in pressure and decreases with increases in temperature.

When surface are increases then physical absorption also increases.

Physical absorption can perform multiple layer of absorption.
3

### JEE Main 2019 (Online) 9th January Morning Slot

Adsorption of a gas follows Freundlich adsorption isotherm. In the given plot, x is the mass of the gas adsorbed on mass m of the adsorbent at pressure P. ${x \over m}$ is proportional to :

A
P2
B
P${^{{1 \over 4}}}$
C
P${^{{1 \over 2}}}$
D
P

## Explanation

According to Freundlich adsorption isotherm, in the median range of pressure

${x \over m} \propto {P^{{1 \over n}}}$

$\Rightarrow$  ${x \over m}$ = kP$^{{1 \over n}}$

taking log both sides, we get,

log${x \over m}$ = logk + ${1 \over n}$ logP

Here in graph between log${x \over m}$ and logP, slope is ${1 \over n}$ and intercepts = log k.

From the given graph,

slope = ${2 \over 4}$ = ${1 \over 2}$

$\therefore\,\,\,$ ${1 \over n}$ = ${1 \over 2}$

$\Rightarrow$   n = 2

$\therefore\,\,\,$ ${x \over m}$ = kP$^{{1 \over 2}}$

$\therefore$${x \over m} \propto {P^{{1 \over 2}}}$
4

### JEE Main 2019 (Online) 9th January Evening Slot

For coagulation of arsenious sulphate sol, which one of the following salt solution will be most effective ?
A
BaCl2
B
AlCl3
C
NaCl
D
Na3PO4

## Explanation

Sulphide is $-$ve charged colloid.

As coagulation is directly proportional to the effective charge, so cation with maximum charge will be most effective for coagulation.

Al3+ > Ba2+ > Na+ coagulating power.