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A $ 0.004\;M$ solution of $K_2SO_4$ is isotonic with a $0.010\;M$ Solution of glucose at the same temperature . Tge apparent per cent degree of dissociation of $K_2SO_4$ is

$\begin{array}{1 1} 25\% \\50\% \\ 75\% \\ 100 \% \end{array} $

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Answer :$ 75\% $
The dissociation of $K_2SO_4$ may be written as
$K_2SO_4 \rightleftharpoons 2K^{+} +SO_4^{2-}$
The concentration of species in the solution is $c(1 - \alpha) +c(2 \alpha)+c \alpha=c(1+2 \alpha)$
Hence $c(1+2 \alpha)= 0.01 \;M$
$\alpha =\large\frac{1}{2}\bigg( \large\frac{0.01 }{0.004 } -\normalsize 1 \bigg) $$=0.75$
The percentage dissociation is $ 0.75 \times 100$ i.e $75 \%$
answered Aug 7, 2014 by meena.p

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