Q. When a quantity of electricity is passed through ${CuSO}_{4}$ solution, 0.16 g of copper gets deposited. If the same quantity of electricity is passed through acidulated water, then the volume of $H_{2}$ liberated at STP will be [Given , atomic weight of Cu = 64]
A) ${4.0\ cm}^{3}$
B) ${56\ cm}^{3}$
C) ${604\ cm}^{3}$
D) ${8.0\ cm}^{3}$
Explanation
(d)
$\frac{Wt.of\ Cu\ deposited}{Wt.of\ H_{2}\ produced} = \ \frac{eq.wt.of\ Cu}{eq.wt.of\ H}$
$\frac{0.16}{wt.of\ H_{2}\ } = \ \frac{64/2}{1} = \ \frac{32}{1}$
Wt. of $H_{2}\ \ = \frac{0.16}{32} = 5\ \times \ 10^{- 3}\ g\$
Volume of $H_{2}$ liberated at STP
$= \ \frac{22400}{2}\ \times 5\ \ \times \ 10^{- 3}cc$
= 56 cc
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.