$0.5 \mathrm{~g}$ mixture of $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ and $\mathrm{KMnO}_{4}$ was…

$0.5 \mathrm{~g}$ mixture of $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ and $\mathrm{KMnO}_{4}$ was
treated with excess of $\mathrm{KI}$ in acidic medium. $\mathrm{I}_{2}$ liberated required $100 \mathrm{~cm}^{3}$ of $0.15 \mathrm{~N} \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}$
solution for titration. The percentage amount of $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ in the mixture is
  1. $85.36 \%$
  2. $14.64 \%$
  3. $58.63 \%$
  4. $26.14 \%$

Solution

Let the amount of the $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ in the mixture be $x \mathrm{~g}$, then amount of $\mathrm{KMnO}_{4}$ will be $(0.5-x) g$ $\therefore \quad\left(\frac{x}{49}+\frac{0.5-x}{31.6}ight)=\frac{100 \times 0.15}{1000}$
where 49 is Eq. wt. of $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ and $31.6$ is
Eq. wt. of $\mathrm{KMnO}_{4}$. On solving, we get $x=0.073 \mathrm{~g}$
Percentage of $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$
$=\frac{0.0732 \times 100}{0.5}=14.64 \%$ ,

Asked in: JEE-TOPICTESTS-CHEMISTRY

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