Question

How many atoms of calcium will be deposited from a solution of $CaC{l}_2$ by a current of $5mA$ flowing for $60$ seconds ?

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

Answer 1

Answer: (A)

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