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STD 12 - 3. Chemical kinetics — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 3. Chemical kinetics4 Marks
MCQ
An organic compound undergoes first order decompostion. The time taken for its decomposition to $\frac{1}{8}$ and $\frac{1}{10}$ of its initial concentration are $t_{1/8}$ and $t_{1/10}$ respectively. What is the value of $\frac{{{t_{1/8}}}}{{{t_{1/10}}}}$ ?   $[\log\, 2 = 0.30]$
  • A
    $0.09$
  • ✓
    $0.9$
  • C
    $9$
  • D
    $90$
✓

Answer

Correct option: B.
$0.9$
b
$\mathrm{kt}_{1 / 8}=2.303 \log \frac{\mathrm{a}}{\mathrm{a} / 8}$

$\mathrm{kt}_{1 / 10}=2.303 \log \frac{\mathrm{a}}{\mathrm{a} / 10}$

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