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0.05 g of a piece of metal in dilute aci...

`0.05 g` of a piece of metal in dilute acid gave `24.62 mL` of `H_(2)` at `27^(@)C` and `760mm` pressure. The `Ew` of metal is

A

25

B

12.5

C

50

D

37.5

Text Solution

Verified by Experts

The correct Answer is:
A
Volume of `H_(2)` at `STP = 24.64 xx (273)/(300) = 22.40 mL`
`22400 mL` of `H_(2)` at `STP = 1 "mole" = 2 "Eqs of" H_(2)`
`22.4 mL "of" H_(2) = (2)/(22400) xx 22.4`
= 0.002 Eqs of `H_(2)`
`= 0.002 "Eq of metal"`
`("Weight")/(Ew) = 0.002`
`(0.05)/(0.002) = Ew = 25`
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