1.

The number of coulombs required to liberate 0.224 dm^(3) of chlorine at 0^(@)C and 1 atm pressure is

Answer»

`2 XX 965`
965/2
965
9650

Solution :`22.4 DM^(3) `= 1 mole of `Cl_(2)`
`THEREFORE 0.224 dm^(3) = 0.01` mole of `Cl_(2)`
As 96,500 COULOMBS `-= 35.5` gm = 0.5 moles of `Cl_(2)`
OR
`(1)/(2)` moles of `Cl_(2) = 96500 implies therefore 0.01` moles
`-= 96500 xx 0.01 xx 2`
`= 965 xx 2` coulombs


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