JEE Main 31-Aug-2021 Shift 2 Solved Paper

According to Freundlich isotherm,

x

m

=kp

1

n

(Using, amount of adsorbate ∝ Volume of absorbate)

10

1

=k×(100)

1

n

...(i)

15

1

=k×(200)

1

n

...(ii)

v

1

=k×(300)

1

n

...(iii)
Divide Eq. (ii) by (i)

15

10

=2

1

x

⇒log(

3

2

)=

1

n

‌log‌2

1

n

=

log‌3−log‌2

log‌2

=

0.4771−0.3010

0.3010

1

n

=0.585
Divide Eq. (iii) by (i)

v

10

=3

1

n

log(

v

10

)=

1

n

‌log‌3
log(

v

10

)=0.585×0.4771=0.2791

v

10

=100.2791
⇒v=10×100.2791
=101.279=10x
x = 1.279
x=128×10−2