43.9 " mL of " `H_2` at STP is produced form 0.1 g of metal
`11200 " mL of " H_2` at STP produced `=(0.1xx11200)/(43.9)`
`=25.5g` of metal
Valency of metal is its lower oxidation state. So
Valency`=(51)/(25.5)=2`
`therefore` Valency`=`Atomic weight/Equivalent weight 0.1 g of the metal in lower oxidation state required 58.8 " mL of " 0.1 `KMnO_4`
51 g of the metal requires `=(58.8xx0.1)/(0.1)xx51`
`=2958 m" Eq of "KMnO_4`
`=2.9 " Eq of "KMnO_4`
`approx3 " Eq of "KMnO_4`
Hence, increase in valency during further oxidation is there. Thus higher oxidation state`=5`
Valency of metal`=2,5`