`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. 6
D. 36

Answer

A

Explanation

1 mol of `O_(2) = 4` eq. of O 22400 mL of `O_(2) 4` eq. of O
`46.6 mL` of `O_(2) = (4)/(22400) xx 46.6 = 0.00832 eq.` Equivalent of metal = Equivalent of O
`("Weight")/("Equivalent weight") = 0.00832 " "(0.1)/(E) = 0.00832 " " :." " E = (0.1)/(0.00832) = 12.0`