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What volume of O(2) at NTP liberated by...

What volume of `O_(2)` at NTP liberated by 5 A current flowing for 193 and through acidulated water ?

A

56 mL

B

112 m L

C

158 mL

D

965 m L

Text Solution

Verified by Experts

The correct Answer is:
A
`W = ( I xx t xx E)/(F) = (5 xx 193 xx 8)/(96500) = 0.08 g `
At NTP , volume of 32 g of `O_(2)` = 22400 mL
`therefore 0.08` g = x
` x = (22400 xx 0.08)/(32) = 56 ` mL
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