What will be the volume of `O_(2)` Liberated at NTP by passing 5 A current For 193 sec. through acidified water.

Options

A. 56 mL
B. 112 mL
C. 158 mL
D. 965 mL

Answer

A

Explanation

`W = (C xx t xx E)/F`
`= (5 xx 193 xx 8)/(96500)`
`because` At NTP, volume of
`32 g O_2 = 22400 mL`
`because ` Volume of `0.08 g O_2 = (22400 xx 0.08)/32`
= 56 mL