A polyvalent metal weghing 0.1 g and having atomic weight 51 reacted will dil `H_2SO_4` to give 43.9 " mL of " `H_2` at STP. This solution containing the metal in the lower oxidation state was found to require 58.8 " mL of " 0.1 permanganate for complete oxidation. What are the valencies of the metal.

Explanation

Suppose the lower oxidation number of the metal is X .
Given that :
` {:("metal "+,H_(2)SO_(4) to ,H_(2)),(0.1 g ,,43.9 " mL at STP"),((0.1)/(51//X)eq,,(43.9)/(11200)eq):}`
(eq . Wt. of the metal = `51/X ` and volume occupied by 1 eq. of hydrogen at NTP = 11200 mL )
Now , eq. of the metal = eq. of hydrogen
`(0.1)/(51//X) = (43.9)/(11200 ) , X = 2 `
Further , the metal is changing from lower oxidation number 2 to higher oxidation number , say , Y .
` :. ` eq. wt of the metal ` = 51/(" change in ON") = 51/(Y-2)`
Eq. of metal = eq. of `KMnO_(4)`
` = (" m.e of " KMnO_(4))/(1000)`
` :. " " (0.1)/(51/(Y-2)) = (0.1 xx 58.8)/1000`
` :. " " Y = 5 ` .