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JEE Mains · Chemistry · STD 12 - 3. Chemical kinetics

An organic compound undergoes first order decomposition. The time taken for decomposition to \(\left(\frac{1}{8}\right)^{\text {th }}\) and \(\left(\frac{1}{10}\right)^{\text {th }}\) of its initial concentration are \( t_{1/8} \) and \( t_{1/10} \) respectively.
What is the value of \( \frac{t_{1/8}}{t_{1/10}} \times 10 \)?
(Given: \( \log 2 = 0.3 \))

  1. A 9
  2. B 0.9
  3. C 3
  4. D 30
Verified Solution

Answer & Solution

Correct Answer

(A) 9

Step-by-step Solution

Detailed explanation

\( t = \frac{1}{k} \ln \frac{A_0}{A_t} \) \( t_{1/8} = \frac{1}{k} \ln \frac{A_0}{A_0/8} = \frac{1}{k} \ln 8 \) \( t_{1/10} = \frac{1}{k} \ln \frac{A_0}{A_0/10} = \frac{1}{k} \ln 10 \) \( \frac{t_{1/8}}{t_{1/10}} = \frac{\ln 8}{\ln 10} = \frac{\log 8}{\log 10}\)…
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