$$0.5\,g$$  mixture of $${K_2}C{r_2}{O_7}$$   and $$KMn{O_4}$$   was treated with excess of $$KI$$  in acidic medium. $${I_2}$$  liberated required $$100\,c{m^3}$$  of $$0.15\,N\,N{a_2}{S_2}{O_3}$$    solution for titration. The percentage amount of $${K_2}C{r_2}{O_7}$$   in the mixture is

Solution :
Let the amount of the $${K_2}C{r_2}{O_7}$$   in the mixture be $$x\,g,$$  then amount of $$KMn{O_4}$$  will be $$\left( {0.5 - x} \right)g$$
$$\eqalign{ & \therefore \,\,\left( {\frac{x}{{49}} + \frac{{0.5 - x}}{{31.6}}} \right) \cr & = \frac{{100 \times 0.15}}{{1000}} \cr} $$
where $$49$$  is $$Eq.\,wt.$$  of $${K_2}C{r_2}{O_7}$$   and $$31.6$$  is $$Eq.\,wt.$$  of $$KMn{O_4}.$$
$$\eqalign{ & {\text{On solving, we get}}\,x = 0.073\,g \cr & {\text{Percentage of}}\,{K_2}C{r_2}{O_7} \cr & = \frac{{0.0732 \times 100}}{{0.5}} \cr & = 14.64\% \cr} $$