Question

For the given electrolyte $A_x{B}_y$, the degree of dissociation '$\alpha$' can be given as:

• A. $\alpha =\frac{i-1}{x+y-1}$
• B. $i=(1-\alpha )+x\alpha +y\alpha$
• C. $a=\frac{1-i}{1-x-y}$
• D. all of these

Answer 1

Answer: (D)

all of these

For the given electrolyte $A_x{B}_y$, the degree of dissociation '$\alpha$' can be given as

$A_x{B}_y\to xA+yB$
1-$\alpha$ $x\alpha$ $y\alpha$

Total moles $=1+(x+y-1)\alpha$

$i=$ moles after dissociation/initial moles $=\frac{+(x+y-1)\alpha }{1}=1+(x+y-1)\alpha$

(A) $\alpha =\frac{i-1}{x+y-1}$

(B) $i=(1-\alpha )+x\alpha +y\alpha$

(C) $a=\frac{1-i}{1-x-y}$

Here, $i$ is the Van't Hoff factor and x and y are the number of cations and anions obtained respectively on the dissociation of 1 molecule of electrolyte.