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Sucrose decomposes by first order kinetics with t1/2 = 3.00 hours. What fraction of the sucrose sample remains after 6 hours?
A1/2
B1/3
C1/4
D1/8
Answer & Solution
Correct answer: C. 1/4
1. For first order, half-life is constant at 3.00 hours.
2. In 6 hours there are $6/3 = 2$ half-lives.
3. Fraction remaining $= \left(\tfrac{1}{2}\right)^2 = \tfrac{1}{4}$.
4. Distractor A is one half-life; distractor D is three half-lives.
_Source: NCERT Class 12 Chemistry Ch 3 "Chemical Kinetics", p.26_
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