A metal oxide has the formula `Z_(2)O_(3)` . It can be reduced by hydrogen to give free metal and water . 0.1596 g of the metal oxide requires 6 mg of hydrogen for complete reduction . The atomic weight of the metal is

Options

A. 27.9
B. 159.6
C. 79.8
D. 55.8

Answer

D

Explanation

The reaction may given as
`Z_(2)O_(3)+3H_(2) to 2Z+3H_(2)O`
0.1596 g of `Z_(2)O_(3)` react with `H_(2)=6mg=0.006 g`
`:.` 1 g of `H_(2) "react with" = (0.1596)/(0.006)=26.6 g "of" Z_(2)O_(3)`
`:.` Eq. wt. of `Z_(2)O_(3)=26.6` (from the definition of eq. wt.)
Eq. wt. of Z +Eq. wt. of O = E + 8 = 26.6
`implies` Eq. wt. of Z = 26.6-8 = 18.6
Valency of metal in `Z_(2)O_(3)=3`
`"Eq. wt. of metal"=("Atomic wt.")/("valency")`
`:.` At. wt. of `Z =18.6 xx 3=55.8`