Question

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

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

Answer 1

Answer: (A)

Key Idea: 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 1.5 C $=\frac{Q×Avogadronumber}{96500}$ $=\frac{1.5×6.023×{10}^{23}}{96500}$ $=9.36×{10}^{18}$ $Ca\to C{a}^{2+}+2e^{−}$ $âˆ\mu$ $2e^{−}$ are required to deposit 1 Ca atom. $∴$ No. of Ca atoms deposited $=\frac{no.ofelectrons}{2}$ $=\frac{9.36×{10}^{18}}{2}$ $=4.68×{10}^{18}$