Question

The degree of dissociation of an electrolyte is $\alpha$ and its van't Hoff factor is $i$. The number of ions obtained by complete dissociation of 1 molecule of the electrolyte is:

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

Answer 1

The correct option is A $\frac{i+\alpha -1}{\alpha }$

Let, $AX$ be a weak electrolyte which dissociation is shown below:.

$AX\stackrel{}{\to }{A}^{+}+X^{-}$

At $t=0$

$\left[AX\right]=1,\left[A^{+}\right]=0,\left[X^{-}\right]=0$

At $t=t^{\prime }$

$\left[AX\right]=1,\left[A^{+}\right]=N\alpha ,\left[X^{-}\right]=1-\alpha$

According to Van't Hoff factor,

$\iota =\frac{1-\alpha +N\alpha }{1}$

$N\alpha =\iota -1+\alpha$

$N=\frac{\iota +\alpha -1}{\alpha }$

this is required solution.