A drop of solution (volume `0.05 mL`) contains `3 xx 10^(-6) "mole" H^(o+)` ions. If the rate constant of disappearance of `H^(o+)` ions is `1 xx 10^(7) mol L^(-1) s^(-1)`, how long would it take for `H^(o+)` ions in the drop of disappear?

Options

A. `6 xx 10^(-8) sec`
B. `6 xx 10^(-7) sec`
C. `6 xx 10^(-9) sec`
D. `6 xx 10^(-10) sec`

Explanation

Since rate constant `=1.0 xx 10^7 mol litre^(-1)sec^(-1)`
`:.` Hence it is zero order reaction.
For zero order
`t = x/k= ("concentration of product")/("rate constant") …(i)`
To calculate concentration :
`:' 0.05 mL has 3xx 10 ^(-6)` mol of `H^(+)`
`:. 1000 mL has (3xx 10^(-6)xx10^3)/0.05= 3/5xx 10^(-1)`
`=0.6 xx 10^(-1) mol //litre" of "H^+`
`:.` By Eq. (i), `t=(0.6xx 10^(-1))/(1xx10^7)=6 xx 10^(-9)`second.