A metal oxide has the formula `X_2O_3`. It can be reduced by hydrogen to give free metal and water. `0.1596g` of metal oxide requires `6mg` of hydrogen for complete reduction. The atomic mass of the metal (in amu) is :

Options

A. 27 .9
B. 159.6
C. 79.8
D. 55.8

Answer

D

Explanation

`Z_(2)O_(3)+3H_(2)to2Z+3H_(2)O`
`:'6xx10^(-3)gH_(2)" reduces"=0.159" g of " Z_(2)O_(3)`
`:.` 1 g `H_(2)` reduces `=(0.1596)/(6xx10^(-3))g Z_(2)O_(3)`
= 26.6 g of `Z_(2)O_(3)`
`:.` Equivalent weight of `Z_(2)O_(3)=26.6`
Equivalent weight of Z+ Equivalent weight of O =26.6
Equivalent weight of Z= (26.6-8)=18.6
Valency of metal in `Z_(2)O_(3)=3`
Equivalent weight `=("Atomic weight")/("Valency")`
Atomic weight `=18.6xx3=55.8`