Question

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. $0.56 \,dm^3$

Answer 1

Answer: (C)

$4.48\,dm^3$

Answer 2

During electrolysis volume of $O _{2}$ and $H _{2}$ liberated are in the ratio of $1: 2$, hence volume of $H _{2}$ liberated will be $4.48 dm ^{3}$.