Hii friends....
Plzz solve 9,10 and 11
✨Plzz don't spam answers
Attachments:
Answers
Answered by
2
It may help you except 9..nehi mila ans.....
Attachments:
Anusha99:
Jo jo pic send karna hai us pic ko click karke rahena.. Aksath.. Ho jayega
Profile pic kaha hai tumhara?
To do
Yup why!?
Hm ok
Answered by
2
Heya ,
Answer of 9th question ,
_______________________
=> The degree of dissociation is calculated by the following relation
α = Λ/ Λ0
Where Λ is the molar conductivity and Λ0 is the limiting molar conductivity of the solution.
=> Now Λ = к/c = [к (Scm-1) X 1000cm3L-1]/ molarity (molL-1)
where к = conductivity of solution and c is the concentration of solution
●Given к = 4.94*10-5 S cm-1 and c = 0.001028 molL-1
=> Therefore,
Λ = [4.94 X 10-5 X 1000] / 0.001028 Scm-2mol-1
= 48.054 Scm-2mol-1
●Given Λ0 = 390.5 Scm-2mol-1
=> therefore α = Λ/ Λ0 = 48.054/390.5 = 0.123
_________________________
Answer of 9th question ,
_______________________
=> The degree of dissociation is calculated by the following relation
α = Λ/ Λ0
Where Λ is the molar conductivity and Λ0 is the limiting molar conductivity of the solution.
=> Now Λ = к/c = [к (Scm-1) X 1000cm3L-1]/ molarity (molL-1)
where к = conductivity of solution and c is the concentration of solution
●Given к = 4.94*10-5 S cm-1 and c = 0.001028 molL-1
=> Therefore,
Λ = [4.94 X 10-5 X 1000] / 0.001028 Scm-2mol-1
= 48.054 Scm-2mol-1
●Given Λ0 = 390.5 Scm-2mol-1
=> therefore α = Λ/ Λ0 = 48.054/390.5 = 0.123
_________________________
Similar questions