Question

How many c.c. of oxygen will be liberated by 2 ampere current flowing for 3 minutes and 13 seconds through acidulated water ?
A. 11.2 c.c.
B. 33.6 c.c.
C. 44.8 c.c.
D. 22.4 c.c.

Answer 1

Correct Answer - D
`W_(O_(2)) = (E xxi xx t)/(F) = (8 xx 2 xx 193)/(96500) = 0.032` g
`(W)/(M) = (V)/(22400) therefore V = (0.032 xx 22400)/(32) = 22.4` c.c.