1.

The volume of water to be added to 100 cm of 0.5 N H_(2)SO_(4)to get decinormal concentration is

Answer»

100 `cm^(3)`
`450 cm^(3)`
`500 cm^(3)`
`400 cm^(3)`

Solution :`N_(1)V_(1) = N_(2)V_(2)`, i.e. `0.5 xx 100 = 0.1 xx V_(2)` or, `V_(2) = 500 cm^(3`.
`THEREFORE` Water to be added to `100 cm^(3)` solution.
`=500 - 100 = 400 cm^(3)`.


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