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0.1 g of metal combines with 46.6 mL of ...

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

A

12

B

24

C

6

D

36

Text Solution

Verified by Experts

The correct Answer is:
A
1 mol of `O_(2) = eq` of `O`
`22400 mL` of `O_(2) = (4)/(22400) xx 46.6`
`= 0.00832 eq`.
Equivalent of metal = Eqivalent of `O`
`("Weight)/(Ew) = 0.00832`
`(0.1)/(0.00832) = 12.0`
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