A certain substance `A` is mixed with an equimolar quantity of substance `B`. At the end of an hour `A` is `75%` reacted. Calculated the time when `A` is `10%` unreacted. (Given order of reaction is zero).

Options

A. `1.2` hours
B. `2.1` hours
C. `3.1` hours
D. `2` hours

Answer

A

Explanation

`C_(t) = C_(0) - kt`
`(C_(0))/(4) = C_(0) - k rArr k = (3)/(4) C_(0)`
Now, `(C_(0))/(10) = C_(0) - (3)/(4)`
`rArr t = (6)/(5) hrs`