1.

When 10 ampere current is passed for 80 min., volume of H_(2) gas liberated is

Answer»

`4.5 dm^(3)`
`5.5 dm^(3)`
`5.57 dm^(3)`
`6.2 dm^(3)`

Solution :QUANTITY of ELECTRICITY =Q=Ampere`xx`Second
`therefore Q=10xx80xx60=48000` COULOMB
`11.2 dm^(3) `of `H_(2)` LIBERATED by 96500 coulomb of electricity. Therefore `x dm^(3)` of `H_(2)` liberated by 4800 coulomb of electricity .
`therefore x=11.2xx4800//96500=5.57`


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