please give the solution to this question
Attachments:
Answers
Answered by
1
Please remember the formula that relates vant hoff factor(i) with the degree of dissociation(α).
i = 1 + (n-1)α
where 'n' is the number of ions present in the solution after dissociation.
For example, in A2B3 the number of particles in the solution after dissociation are 5 namely 2As and 3Bs. So, n=5
In your question the formula is AxBy, therefore there will be xAs and yBs in the solution after dissociation.
Hence here n = x + y
Now substitute it in the above equation, we get
i = 1 + (x + y - 1)α
i - 1 = (x + y -1)α
Therefore α = i - 1/(x + y-1)
i = 1 + (n-1)α
where 'n' is the number of ions present in the solution after dissociation.
For example, in A2B3 the number of particles in the solution after dissociation are 5 namely 2As and 3Bs. So, n=5
In your question the formula is AxBy, therefore there will be xAs and yBs in the solution after dissociation.
Hence here n = x + y
Now substitute it in the above equation, we get
i = 1 + (x + y - 1)α
i - 1 = (x + y -1)α
Therefore α = i - 1/(x + y-1)
Similar questions