Question

The half-life period of a. substance in a certain enzyme catalysed reaction is 138 s. The time required for the concentration of the substance to fall from 1.28 mol^–1 to 0.04 mg L^–1 is

पर्याय

• 4145

• 5528

• 6905

• 2765

Answer 1

6905

Explanation:

Amount left after n half-life period `[A]_0 = [A]_0/2^n`

= `[A]_0/[A] = 1.28  "mg" (1.28  "mg L"^-1)/(0.04  "mg  L"^-1)` 

[2]ⁿ = 82 = [2]⁵

No. of tY2 = peroids = 5

Time required = 138 x 5 = 6905.