The degree of ionization (.α .) of a weak electrolyte AxBy is related to van't Hoff factor by expression

Question

The degree of ionization (.α .) of a weak electrolyte AxBy is related to van't Hoff factor by expression (A) α =(i - 1/x + y + 1) (B) α =(i - 1/x + y - 1) (C) α =(x + y - 1/i - 1) (D) α =(x + y + 1/i - 1)

Tardigrade Admin · Accepted Answer

α =(i - 1/n - 1) Here, n = number of ions formed = x + y So α =(i - 1/x + y - 1) .

Q. The degree of ionization $(α)$ of a weak electrolyte $A_{x} B_{y}$ is related to van't Hoff factor by expression

$α = \frac{i - 1}{x + y + 1}$

$α = \frac{i - 1}{x + y - 1}$

$α = \frac{x + y - 1}{i - 1}$

$α = \frac{x + y + 1}{i - 1}$

$α = \frac{i - 1}{n - 1}$
Here, n = number of ions formed = x + y
So $α = \frac{i - 1}{x + y - 1}$ .

Q. The degree of ionization (α) of a weak electrolyte Ax​By​ is related to van't Hoff factor by expression

α=x+y+1i−1​

α=x+y−1i−1​

α=i−1x+y−1​

α=i−1x+y+1​

α=n−1i−1​ Here, n = number of ions formed = x + y So α=x+y−1i−1​ .