Question

The number of atoms of $Ca$ that will be deposited from a solution of $CaC{l}_2$ by a current of $25mA$ for $60s$ will be:

• A. $4.68\times {10}^{18}$
• B. $4.68\times {10}^{15}$
• C. $4.68\times {10}^{10}$
• D. $2.34\times {10}^{15}$

Answer 1

The correct option is A $4.68\times {10}^{18}$

$25\times {10}^{-3}A\times 60sec$ of charge carried by

$=\frac{25\times {10}^{-3}A60s}{96500C}=1.55\times {10}^{-5}mol{e}^{-}$

$C{a}^{2+}+2e^{-}\to Ca$

2 mol of $e^{-}$ will produce = $6\times {10}^{23}$ atoms of Ca

$1.55\times {10}^{-5}mol{e}^{-}$ will produce

$=\frac{1.55\times {10}^{-5}\times 6\times {10}^{23}}{2}=4.68\times {10}^{18}$ atoms of Ca

Hence, option $A$ is correct.