Question

An organic compound undergoes first-order decomposition. The time taken for its decomposition to 1/8 and 1/10 of its initial concentration are $t_{1/8}$ and $t_{1/10}$ respectively. What is the value of $\frac{t_{1/8}}{t_{1/10}}\times 10$? (take $1o{g}_{10}2=0.3$)

• A. 9
• B. 8
• C. 10
• D. 15

Answer 1

Answer: (A)

9

$t_{\frac{1}{8}}=\frac{2.303\times log8}{k}=\frac{2.303\times 3log2}{k}$ Note : $[log\frac{1}{1/8}=log8]$
$t_{\frac{1}{10}}=\frac{2.303\times log10}{k}=\frac{2.303}{k}$
$\frac{t_{\frac{1}{8}}}{t_{\frac{1}{10}}}\times 10=\frac{\frac{2.303\times 3log2}{k}}{\frac{2.303}{k}}\times 10=9$