How many hour are required for a current of `3.0` ampere to decompose `18 g` water?

Options

A. 9 hrs
B. 12 hrs
C. 18 hrs
D. 24 hrs

Answer

C

Explanation

`H_(2) O hArr H_(2) + (1)/(2) O_(2)`
`therefore` 18 gms (1 mole) liberates 2gms.of `H_(2)` (M.mass of `H_(2) = 2`)
`E = ("At.mass")/("valency") = (1)/(1) implies therefore Z = (1)/(96500)`
t = ? , t = 3 A , W = 2gm .
`t = (W xx 96500)/(E xxi) = (2 xx 96500)/(3) 64332 "sec" = (64322)/(3600) ` hrs
= 17.9 hrs = 18 hrs .