Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  JEEMAIN and AIPMT  >>  Chemistry  >>  Solutions
0 votes

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} $

Can you answer this question?

1 Answer

0 votes
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

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App