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The number of atoms of Ca that will be d...

The number of atoms of `Ca` that will be deposited from a solution of `CaCl_(2)` by a current of `25mA` for `60s` will be

A

`4.68xx10^(18)`

B

`4.68xx10^(15)`

C

`4.68xx10^(10)`

D

`2.34xx10^(15)`

Text Solution

Verified by Experts

The correct Answer is:
a
`25xx10^(-3)Axx60 C` of charge carried by
`=(25xx10^(-3)xx60s)/(96500C)=1.55xx10^(-5) mol e^(-)`
`Ca^(2+)+2e^(-) rarr Ca`
`2 mol ` of `e^(-)` will produce `=6xx10^(23)` atoms of `Ca`
`1.55xx10^(-5) mol e^(-)` will produce
`=(1.55xx10^(-5)xx6xx10^(23)/(2)=4.68xx10^(18)` atoms of `Ca`
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