Question

The relation between Vant Hoff factor (1) and the degree of dissociation ( $\alpha$ ) for a weak electrolite given by formula AxBy is:

• A. $\alpha =\frac{i-1}{x+y-1}$
  
• B. $\alpha =\frac{i-1}{x+y+1}$
  
• C. $\alpha =\frac{x+y-1}{i-1}$
  
• D. $\alpha =\frac{x+y+1}{i-1}$

Answer 1

The correct option is A

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

$AxBy\to xA+yBInitialConc.:C00finalConc:C(1-\alpha )xC\alpha yC\alpha Now,i=\frac{C(1-\alpha )+xC\alpha +yC\alpha }{C\alpha }=1-\alpha +x\alpha +y\alpha \therefore \alpha =\frac{i-1}{(x+y-1)}$