Question

Relationship between van't Hoff's factor (i) and degree of dissociation (α) is ______.

Options

• i = `(alpha - 1)/("n'" - 1)`

• i = `(alpha - 1)/(1 - "n'")`

• `alpha = (1 - "i")/("n'" - 1)`

• `alpha = ("i" - 1)/("n'" - 1)`

Answer 1

Relationship between van't Hoff's factor (i) and degree of dissociation (α) is `underline(alpha = ("i" - 1)/("n'" - 1))`.

Explanation:

Relationship between vant Hoff factor (i) and degree of dissociation (α) is given by

`alpha = ("i" - 1)/("n'" - 1)`

where n is the number of ions formed after dissociation.

The relationship can be obtained as follows:

For the reaction, \[\ce{A <=> \text{n' B}}\]

Initially 1 mole 0

After dissociation (1 - α) mole n' α

Total number of moles present in the solution

= (1 - α) + n' α = 1 + (n' - 1)α

van't Hoff factor, i = 1 + (n' - 1), α > 1 if n' ≥ 2

`therefore alpha = ("i" - 1)/("n'" - 1)`