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How many c.c. of oxygen will be liberate...

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.

Text Solution

Verified by Experts

The correct Answer is:
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.
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