1.

How many c.c. of oxygen will be liberated by 2 ampere current flowing for 3 minutes and 13 seconds through acidulated water ?

Answer»

11.2 c.c.
33.6 c.c.
44.8 c.c.
22.4 c.c.

Solution :`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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