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For a first order reaction Half-life period, t1/2 = 0.693/k = 0.693/2.303 × 10^-3 s^-1 = 300.91 s Now, 40 g → 20 g → 10 g So, 40 g substance requires 2 half-life periods to reduce upto 10 g. ∴ Time taken in reduction = 2 × 300.91 s = 601.82 ≈ 602 s

1. Half-life period, t1/2 = 0.693/k
2. t1/2 = 0.693/2.303 × 10^-3 s^-1
3. Time taken in reduction = 2 × 300.91 s
4. Time taken in reduction = 601.82 ≈ 602 s

Correct answer: Time taken in reduction = 601.82 ≈ 602 s

Solution

The time taken for the reduction of 40 g to 10 g is correctly calculated as 2 half-life periods, which equals approximately 602 seconds.

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