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An organic compound decomposes by first-order kinetics. Let t(1/8) and t(1/10) be the times for the concentration to fall to 1/8 and 1/10 of its initial value respectively. Find the value of [t(1/8) / t(1/10)] * 10. (Take log10(2) = 0.3.)
- 9
- 8
- 10
- 7
Correct answer: 9
Solution
The ratio t(1/8)/t(1/10) = log10(8)/log10(10) = 0.9/1 = 0.9, so multiplying by 10 gives 9.
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