The decompoistion of `H_(2)O_(2)` can be followed by titration with `KMnO_(4)` and is found to be a first order reaction. The rate constant is `4.5 xx 10^(-2)`. In an experiment, the initial titre value as `25 mL`. The titre value will be `5 mL` after a lapse of

Options

A. `4.5 xx 10^(-2) xx 5 min`
B. `(log_(e) 5)/(4.5 xx 10^(-2)) min`
C. `(log_(e)5//4)/(4.5 xx 10^(-2))min`
D. None of these

Answer

B

Explanation

Use the formula
`t = (2.303)/(k)log.(V_(0))/(V_(t))`
`= (1)/(k)ln.(V_(0))/(V_(t))`
`= (1)/(4.5 xx 10^(-2) min^(-1))ln.(25mL)/(5 mL)`
`= (log_(e)5)/(4.5 xx 10^(-2))min`