The number of coulombs required to liberate 0.224 `dm^(3)` of chlorine at `0^(@)C` and 1 atm pressure is
A. `2 xx 965`
B. 965/2
C. 965
D. 9650

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1 Answers

Correct Answer - A
`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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