$5\times {10}^{-5}$ moles of $KMn{O}_4$ in acidic medium is completely consumed to oxidise $5\times {10}^{-5}$ moles of $X^{n+}\to X{O}_4^{-}$ then what is the value of $2n ?$

 $5\times {10}^{-5}moleofKMn{O}_4/H^{+}+5\times {10}^{-5}molesof{x}^{+n}\to X{O}_4^{-}$
 $5\times {10}^{-5}\times 5=5\times {10}^{-5}(7-n)$
 $25\times {10}^{-5}=35\times {10}^{-5}-5n{10}^{-5}$
 $25\times {10}^{-5}={10}^{-5}[35-5n]$
 $5n=35-25$
 $5n=10$
 $n=2\stackrel{ }{\to }2nis4$