JEE Chemistry 2026 Question with Solution

Given: An organic compound undergoes first order decomposition. The times are $t_{1/8}$ and $t_{1/10}$ when concentration becomes $\dfrac{1}{8}$ and $\dfrac{1}{10}$ of the initial concentration respectively.

Find: $\dfrac{t_{1/8}}{t_{1/10}}\times 10$

For a first order reaction,

$$t = \frac{2.303}{k}\log\left(\frac{[A]_0}{[A]}\right)$$

For concentration becoming $\dfrac{1}{8}$ of the initial value,

$$t_{1/8}=\frac{2.303}{k}\log 8$$

Using $\log 8 = 3\log 2 = 3 \times 0.3 = 0.9$,

$$t_{1/8}=\frac{2.303}{k}\times 0.9$$

For concentration becoming $\dfrac{1}{10}$ of the initial value,

$$t_{1/10}=\frac{2.303}{k}\log 10 = \frac{2.303}{k}$$

Now,

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

Therefore,

$$\frac{t_{1/8}}{t_{1/10}}\times 10 = 0.9 \times 10 = 9$$

The solution working gives $9$, so the correct option is C. The solution also shows a contradictory boxed value of $3$, but that does not match the calculation.