Q. How many atoms of calcium will be deposited from a solution of ${CaCl}_{2}$ by a current 0.25 mA flowing for 60 s?
A) $4.68\ \times 10^{18}$
B) $2.34\ \times 10^{18}$
C) $1.24\ \times 10^{18}$
D) $0.46\ \times 10^{18}$
Explanation
(a)
Given, current $(i) = 0.25\ mA = 0.025\ A$
Time $(t) = 60\ s$
$Q = i \times t = 60 \times 0.025 = 1.5\ C$
$No.of\ electrons = \frac{1.5 \times 6.023 \times 10^{23}}{96500}$
$e^{-} = 9.36 \times 10^{18}$
$Ca \rightarrow {Ca}^{2 +} + 2e^{-}$
$2e^{-}$ are required to deposite one Ca atom
$9.36 \times 10^{18}\ e^{-}$ will be used to deposite
$= \frac{9.36 \times 10^{18}}{2} = 4.68\ \times 10^{18}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.