`2.68xx10^(-3)` moles of solution containing anion `A^(n+)` require `1.61xx10^(-3)` moles of `MnO_(4)^(-)` for oxidation of `A^(n+)` to `AO_(3)^(-)` in acidic medium. What is the value of `n`?

Options

A. 1
B. 2
C. 3
D. 4

Answer

B

Explanation

The reaction are
`MnO_(4)^(-) + 8 H^(+) + 5e^(-) rarr Mn^(+2) + 4H_(2)O`
`A^(+n) + 3H_(2)O rarr AO_(3)^(-) + 6H^(+) + ( 5-n) e^(-)`
Amount of electrons involved in the given amount of `MnO_(4)^(-) = 5 xx 1.6 xx 10^(-3) `mol.
Equating these two we get `5 xx 1.6 xx 10^(-3)`
`= ( 5-n) 2.7 xx 10^(-3) `
`:. n = 2 ` ( approx. )