When an electric current is passed through acidified water, `112ml` of `H_(2)` gas at `NTP` is collected at the cathode is `965` seconds. The current passed in amperes is

Options

A. `(2.0)`
B. `(1.0)`
C. `(0.1)`
D. `(0.5)`

Explanation

`2H^(+)2e^(-e)toH_2` ( at cathode)
`W=ZIt=(E It)/(96500)`
Moles of `H_2` deposited `(W)=" Moles" xx "molar mass"(112)/(22400)xx2`
Thus,`112/22400xx2=(1xxIxx965)/96500=I=1A`