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When a current of 0.75 A is passed throu...

When a current of 0.75 A is passed through a `CuSO_(4)` solution for 25 minutes, 0.369 g of copper is deposited at the cathode. Calculate the atomic mass of copper.

Text Solution

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`Cu^(+)(aq)+underset(2" moles")(2e^(-)) to underset(1" mole")(Cu(s))`
Quantity of charge passed `=lxxt=(0.75A)xx(25xx60s)`
`=1125 A .S.=1125" C"`
`1125" C"` of charge deposit copper=0.369 g
`2xx96500" C mol"^(-1)` of charge deposit `=((0.369g))/((1125C))xx(2xx96500 " C mol"^(-1))=63.3" g mol"^(-1)`.
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