Derive the equation, `alpha=(i-1)/(n-1)`
Derive a relation between degree of dissociation and van't Hoff factor.

Explanation

Consider one mole of a substance An (like an electrolyte), which on dissociation in the solution gives n number of particles A. Let `alpha` be the degree of dissociation at equilibrium. Then, the dissociation equilibrium can be represented as,
`An Leftrightarrow nA`
`{:(,"At start",1,0,"mole"),(,"At equilibrium",1-alpha,nalpha,"mole"):} `
At start, let the number of moles of An (or particles) be 1.
At equilibrium
The total number of moles of particles `=1-alpha+nalpha`
`=1+nalpha-alpha`
`=1+alpha(n-1)`
Theoretical colligative properties are due to 1 mole while the observed colligative properties are due to `1+alpha(n-1)` moles of particles in the solution.
The van't Hoff factor i, will be,
`i=("Observed colligative property")/("Theoretical colligative property")`
Now,
Colligative property `alpha` Number of particels in the solution.
` therefore i=(1+alpha(n-1))/(1)`
`therefore hati=1+alpha(n-1)`
`therefore alpha(n-1)=hati-1`
`therefore alpha=(i-1)/((n-1))`
This is the relation between degree of dissociation and van't Hoff factor i.