If the half life of a first order reaction is 60 min, the approximate time in min, required to complete 90% of the reaction is (log2 = 0.3) 

Time Required to Complete 90% of a First Order Reaction

For a first-order reaction, the time required to complete 90% of the reaction (i.e., when 10% of the reactant remains) can be calculated using the relationship between half-life and the reaction's rate constant. The half-life (t1/2) is given as 60 minutes, and we use the approximation log10 2 = 0.3 as specified.

1. Rate Constant (k)

k = ln 2 / t1/2 = ln 2 / 60
Since ln 2 = log10 2 × ln 10 and log10 2 = 0.3,
we approximate ln 2 ≈ 0.3 × ln 10.
 

2. Time for 90% Completion (t90)

For first-order kinetics:
t = ln([A]0/[A]) / k, where [A]0/[A] = 10 (10% remains).
So,
t90 = ln 10 / k = (ln 10 / ln 2) × t1/2
Using ln 10 / ln 2 = log2 10 = 1 / log10 2 = 1 / 0.3:
t90 = (1 / 0.3) × 60 = 200 minutes
 

Result:
The approximate time required to complete 90% of the reaction is 200 minutes. 
This aligns with the approximation log10 2 = 0.3 and leverages the half-life relationship for first-order kinetics.