Chemistry, asked by imankhan2837, 1 year ago

Conductivity of 0.00241 m acetic acid is 7.896 105s cm1. Calculate its molarconductivity and if omfor acetic acid is 390.5 s cm2mol1, what is it

Answers

Answered by Anonymous
4

HEY MATE ✌️✌️

Given that ,

κ= 7.896 × 10−5S m−1

C=M= 0.00241 molL−1

Theformula ofmolar conductivity,

Λm= (k× 1000)/M

Plug the value we get

Λm= (7.896 × 10−5×1000)/0.00241

= 32.76S cm2mol−1

Theformula ofdegree of dissociation

α= Λm/ Λom

Plug the value we get

α= 32.76S/390.5= 0.084

Theformula ofdissociation constant

K = Cα/(1 – α)

Plug the values we get

K = 0.00241 × 0.084/(1– 0.084)= 1.86 × 10−5mol L−1

PLZZ MARK AS BRAINLIEST ❤️❤️

DO FOLLOW ME

Answered by Anonymous
1

Explanation:

Given: K=7.896×10

−5

Scm

−1

C=M=0.00241molL

−1

∴ Molar conductivity

∧m=(K×1000)/M=(7.896×10

−5

×1000)/0.00241

=32.76Scm

2

mol

−1

Degree of dissociation

α=

λ

o

m

λm

=

390.5

32.76

=0.084

∴ Dissociation constant

⇒K=

1−α

=

(1−0.084)

0.00241×0.084

=1.86×10

−5

molL

−1

Similar questions