`1.520 g` of the hydroxide of a metal on ignition gave `0.995 g` of oxide. The equivalent weight of metal is

Options

A. `1.52`
B. `0.995`
C. `190`
D. `9`

Answer

D

Explanation

`("Wt. of metalhydroxide")/("Wt. of metal oxide")`
`("Eq. wt. of metal hydroxide")/("Eq. wt. of metal oxide")`
`:. (1.520)/(0.995) = (E_(m)+17)/(E_(m)+8)`
or 1.528 `= (E_(m)+17)/(E_(m)+8)`
or 0.528 `E_(m) = 4.78 or E_(m) = 9`