6 × 10^–3 mole K₂Cr₂O₇ reacts completely with 9 × 10^–3 mole X⁺ⁿ to give ${\mathrm{XO}}_3^{-}$ and Cr⁺³. The value of ‘n’ is

$$\begin{gathered} {\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7+6{\mathrm{e}}^{-}+14{\mathrm{H}}^{+}\to 2{\mathrm{Cr}}^{+3}+7{\mathrm{H}}_2\mathrm{O} \\ {\mathrm{X}}^{+\mathrm{n}}+3{\mathrm{H}}_2\mathrm{O}\to {\mathrm{XO}}_3^{-}+(5-\mathrm{n}){\mathrm{e}}^{-}+6{\mathrm{H}}^{+} \\ \mathrm{given}, {\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7 : {\mathrm{X}}^{+\mathrm{n}}=6\times {10}^{-3} : 9\times {10}^{-3}=6:9 \\ \mathrm{From} \mathrm{eq}. {\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7 : {\mathrm{X}}^{+\mathrm{n}}=(5-\mathrm{n}):6 \\ \therefore \frac{(5-\mathrm{n})}{6}=\frac{6}{9} \\ \mathrm{n}=1 \end{gathered}$$