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
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
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
Cα
=
(1−0.084)
0.00241×0.084
=1.86×10
−5
molL
−1
Similar questions