A metal oxide has the fomula `M_2O_3`. It can be reduced by hydrogen to give free metal and water. 0.16g of the metal requires 6mg of `H_2` for complete reduction. The atomic mass of the metal is .n? times the atomic number of calcium. The value of .n. is

Answer

D

Explanation

`M_(2) O_3 +3H_(2(g)) rarr 2M +3H_2O`
6g gives`H_2` 2 moles of metal
`:. 6xx10^(-3)` gm of `H_2` give 0.002 moles of .M.
`0.002 = (0.16)/("M.Wt")`
`:.` M.Wt `=(0.16)/(0.002)=80`