Question

0.5 g sample of iron containing mineral mainly in the form of $CuFe{S}_2$ was reduced suitably to convert all the ferric ions into ferrous ions $(F{e}^{3+}⟶F{e}^{2+})$ and was obtained as solution. In the absence of any interfering radical, the solution required 42 mL of 0.01 M $K_{2}C{r}_2{O}_7$ for titration. Calculate $\%$ of $CuFe{S}_2$ in sample.

• A. $92.48\%$
• B. $82.48\%$
• C. $72.48\%$
• D. $62.48\%$

Answer 1

The correct option is A $92.48\%$

The number of moles of potassium dichromate $=\frac{42\times 0.01}{1000}=0.00042$

1 mole of potassium dichromate oxidizes 6 moles of ferrous ions.

Hence, the number of ferrous ions $=6\times 0.00042=0.00252$ mol

The mass of $CuFe{S}_2=183.6\times 0.00252=0.4626g$

The percentage purity of the sample $=\frac{0.4626\times 100}{0.5}=92.48$