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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

18

D

36

Text Solution

Verified by Experts

The correct Answer is:
A
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`
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