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The degree of dissociation $'\alpha'$ of a weak electrolyte is (N is number of ions given by 1 mol of the electrolyte)

$\begin{array}{1 1}(a)\;\large\frac{i-1}{N-1}\\(b)\;\large\frac{i-1}{N+1}\\(c)\;\large\frac{N+1}{i-1}\\(d)\;\large\frac{N-1}{i-1}\end{array}$

1 Answer

Let, AX be a weak electrolyte whose dissociation is shown below:.
$AX\rightarrow A^++X^-$
At t=0, [AX]=1, [$A^+$]=0, and [$X^-$]=0
At t=t', [AX]=1, [$A^+$]=N$\alpha$, and [$X^-$]=1-$\alpha$
Van't Hoff factor, $i=\large\frac{1-\alpha +N\alpha}{1}$
$i=1-\alpha[1-N]$
$\Rightarrow \alpha [1-N]=1-i$
$\alpha =\large\frac{1-i}{1-N}$
$\Rightarrow \large\frac{i-1}{N-1}$
Hence (a) is the correct answer.
answered Feb 14, 2014 by sreemathi.v
edited Jul 20, 2014 by mosymeow_1
 
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