Question

How many atoms of calcium will be deposited from a solution of $ CaCl_{2} $ by a current of $5\, mA$ flowing for $60$ seconds ?

• A. $ 4.68\times {{10}^{18}} $
• B. $ 4.68\times {{10}^{15}} $
• C. $ 4.68\times {{10}^{12}} $
• D. $ 4.68\times {{10}^{9}} $

Answer 1

Answer: (A)

First calculate total current passed $(Q)$ by equating it to as follows
$Q=i t$
then number of electrons and then number of atoms of Ca deposited.
Given, $i=25\, mA $
$=0.0025 \,A$
$t=60\, s$
$Q=i t$
$=0.0025 \times 60$
$=1.5\, C$
Number of electrons in $1.5 \,C$
$=\frac{Q \times \text { Avogadro number }}{96500}$
$=\frac{1.5 \times 6.023 \times 10^{23}}{96500}$
$=9.36 \times 10^{18}$
$Ca \rightarrow Ca ^{2+}+2 e^{-}$
$\because 2 e^{-}$are required to deposit $1\, Ca$ atom.
$\therefore $ no. of $Ca$ atoms deposited
$=\frac{\text { no. of electrons }}{2}$
$=\frac{9.36 \times 10^{18}}{2}$
$=4.68 \times 10^{18}$