• # question_answer 30) $0.5\text{ }g$ 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 $100c{{m}^{3}}$ of $0.15N$. $N{{a}_{2}}{{S}_{2}}{{O}_{3}}$ solution for titration. The percentage amount of ${{K}_{2}}C{{r}_{2}}{{O}_{7}}$ in the mixture is A)  $85.36%$                   B)  $14.64%$C)  $58.63%$                   D)  $26.14%$

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 $(0.5-x)g$ $\therefore$   $\left( \frac{x}{49}+\frac{0.5-x}{31.6} \right)=\frac{100\times 0.15}{1000}$ where 49 is Eq. wt. of ${{K}_{2}}C{{r}_{2}}{{O}_{7}}$ and $3.16$ is Eq. wt. of $KMn{{O}_{4}}$. On solving, we get $x=0.073\text{ }g$ % age of ${{K}_{2}}C{{r}_{2}}{{O}_{7}}=\frac{0.0732\times 100}{0.5}=14.64%$