0.5 g mixture of ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$ and ${\mathrm{KMnO}}_4$ was treated with excess of KI in acidic medium Iodine liberated required $150 {\mathrm{cm}}^3$ of  $0.10\mathrm{N}$ solution of thiosulphate solution for titration. Then % of  ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$ is ________

( MW: K₂Cr₂O7= 294 , KMnO₄= 158)

Equivalent weight of $K_{2}C{r}_2{O}_7=\frac{294}{6}=49$

Equivalent weight of $KMn{O}_4=\frac{158}{5}=31.6$

            m  eq of $K_{2}C{r}_2{O}_7+$ m eq of$KMn{O}_4=$ m eq of $I_2$= m eq of hypo

            let the mass of $K_{2}C{r}_2{O}_7$= $xg$

            the mass of $KMnO4 = (0.5-x)g$

$\frac{x}{49}+\frac{0.5-x}{31.6}=150\times {10}^{-3}\times 0.1$

$x=0.0732$

$\%K_{2}C{r}_2{O}_7=\frac{0.0732}{0.5}\times 100$

=14.64