An organic compound undergoes first order decomposition. The time taken for decomposition to (1/8)^(th) and (1/10)^(th) of its initial concentration are t 1/8 and t 1/10 respectively. What is the value of (t 1/8)/(t 1/10)×10 ? (log 2 = 0.3)

Chemical Kinetics: JEE Main 2026 Chemistry question

JEE Main 2026 · 28 Jan, Morning shift · Q53

JEE Main 202628 Jan, Morning shiftChemistryMediumChemical Kinetics62× since 2002

An organic compound undergoes first order decomposition. The time taken for decomposition to

\left(\frac18\right)^{\text{th}}

and

\left(\frac1{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}}\times10

? (log 2 = 0.3)

JEE Main 2026 Chemical Kinetics Question with Solution (28 Jan Morning shift, Q53) | JEE Arc