Q. During electrolysis of water the volume of $O_{2}$ liberated is 2.24${dm}^{3}$. The volume of hydrogen liberated, under same conditions will be
A) ${2.24\ dm}^{3}$
B) ${1.12\ dm}^{3}$
C) ${4.48\ dm}^{3}$
D) ${0.56\ dm}^{3}$
Explanation
(c)
During electrolysis, volumes of $O_{2}$ and $H_{2}$ liberated are in the ratio of 1 : 2 Hence, volume of $H_{2}$ liberated will be $4.48{dm}^{3}.$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.