Question

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

• 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)

Given, $i=25\, m A=0.0025\, A$
$t=60\, s $
$Q=$ it
$=0.0025 \times 60 $
$=1.5\, C$
number of electrons in $1.5\, C$
$=\frac{Q \times \text { Avogadro number }}{96500}$
$=\frac{1.5 \times 6.023 \times 10^{23}}{96500}$
$=9.36 \times 10^{18}$
$Ca \rightarrow Ca ^{2+}+2 e^{-}$
$2 e^{-}$ are required to deposit $1\, Ca$ atom.
$\therefore $ number of $Ca$ atoms deposited
$=\frac{\text { no. of electrons }}{2}$
$=\frac{9.36 \times 10^{18}}{2}$
$=4.68 \times 10^{18}$