The number of atoms of `Ca` that will be deposited from a solution of `CaCl_(2)` by a current of `25mA` for `60s` will beA. `4.68xx10^(18)`B. `4.68xx10^(15)`C. `4.68xx10^(10)`D. `2.34xx10^(15)`

The number of atoms of `Ca` that will be deposited from a solution of

1.	The number of atoms of `Ca` that will be deposited from a solution of `CaCl_(2)` by a current of `25mA` for `60s` will beA. `4.68xx10^(18)`B. `4.68xx10^(15)`C. `4.68xx10^(10)`D. `2.34xx10^(15)`
Answer» Correct Answer - 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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The number of atoms of `Ca` that will be deposited from a solution of `CaCl_(2)` by a current of `25mA` for `60s` will beA. `4.68xx10^(18)`B. `4.68xx10^(15)`C. `4.68xx10^(10)`D. `2.34xx10^(15)`

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