Q. By passing 9.65 A current for 16 min 40 s , the volume of $O_{2}$ liberated at STP will be
A) 280 mL
B) 560 mL
C) 1120 mL
D) 2240 mL
Explanation
(c)
$2 \times \ 96500\ C$ electricity is used to liberate
$= \ 22400\ mL\ O_{2}\ at\ STP\ \$
∴ $9.65\ \times \ 1000\ C$ electricity will liberate
$= \ \frac{22400\ \times \ \ 9.65\ \times 1000}{2\ \times 96500}$
= 1120mL
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.