How many atoms of calcium will be deposited from a solution of `CaCl_(2)` by a current of 5mA flowig for 60s?

Options

A. `4.68xx10^(18)`
B. `4.68xx10^(15)`
C. `4.68xx10^(12)`
D. `4.68xx10^(9)`

Answer

A

Explanation

Givenn `i=25mA=0.0025A`
`t=60s`
Q=it
`=0.0025xx60`
`=1.5C`
number of electrons in `1.5C`
`=(Qxx"Avogadro number")/(96500)`
`=(1.5xx6.023xx10^(23))/(96500)`
`=9.36xx10^(18)`
`CatoCa^(2+)+2e^(-)`
`2e^(-)` are required to deposit 1 Ca atom.
`therefore`Number of Ca atoms deposited
`=("no. of electrons")/(2)`
`=(9.36xx10^(18))/(2)`
`=4.68xx10^(18)`