`0.1 g` of metal combines with `46.6 mL` of oxygen at `STP`. The equivalent weight of metal is

Options

A. 12
B. 24
C. 18
D. 36

Answer

A

Explanation

1 mole of `O _(2) = 4 eq.` Of oxygen
`22400 ml of O _(2) = (4)/(22400) xx 46.6`
`= 0. 00832 eq.`
Equivalent of metal = Equivalent of oxygen
`("Weight")/("Equivalent") = 0.00832`
`(0.1)/(E) = 0.00832`
`therefore E = (0.1)/(0.00832) = 12.0`