An organic compound undergoes first order decomposition. The time taken for decomposition to and of its initial concentration are ? and respectively.

Question

An organic compound undergoes first order decomposition. The time taken for decomposition to ${((\frac{1}{8}))}^{th}$ and ${((\frac{1}{10}))}^{th}$ of its initial concentration are $t_{1 / 8}$ and $t_{1 / 10}$ respectively.

What is the value of $\frac{t_{1 / 8}}{t_{1 / 10}} \times 10$

$(\log 2 = 0.3)$ [2026]

What is the value of $\frac{t_{1 / 8}}{t_{1 / 10}} \times 10 ?$

$(\log 2 = 0.3)$ [2026]

Smart Achievers · Accepted Answer

(1)

$t = \frac{1}{k} \ln \frac{A_{0}}{A_{t}}$

$t_{1 / 8} = \frac{1}{k} \ln \frac{A_{0}}{A_{0} / 8} = \frac{1}{k} \ln 8$

$t_{1 / 10} = \frac{1}{k} \ln \frac{A_{0}}{A_{0} / 10} = \frac{1}{k} \ln 10$

$\frac{t_{1 / 8}}{t_{1 / 10}} = \frac{\ln 8}{\ln 10} = \frac{\log 8}{\log 10}$

$\frac{t_{1 / 8}}{t_{1 / 10}} = \log 8 = 3 \log 2 = 0.9$

$\frac{t_{1 / 8}}{t_{1 / 10}} \times 10 = 9$

Solution

1 9

2 0.9

3 3

4 30

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Solution

Q. An organic compound undergoes first order decomposition. The time taken for decomposition to (18)th and (110)th of its initial concentration are t1/8 and t1/10 respectively. What is the value of t1/8t1/10×10 (log2=0.3) [2026]

1 9

2 0.9

3 3

4 30

Ans. (1) t=1klnA0At t1/8=1klnA0A0/8=1kln8 t1/10=1klnA0A0/10=1kln10 t1/8t1/10=ln8ln10=log8log10 t1/8t1/10=log8=3log2=0.9 t1/8t1/10×10=9

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