MHTCET 11 Apr 2026 Shift 2 Paper

Concept:For a first order reaction, the rate constant $k$ is related to half life $t_{1/2}$ by $k = \frac{0.693}{t_{1/2}}$, and the concentration at any time $t$ is given by $[A] = [A]_0 e^{-kt}$.Explanation:First, calculate the rate constant $k$ using the given half life $t_{1/2} = 850 \, \mathrm{s}$.$k = \frac{0.693}{850}$$k = 8.15 \times 10^{-4} \, \mathrm{s}^{-1}$Now, use the first order concentration equation:$[A] = [A]_0 e^{-kt}$Here, $[A]_0 = 0.06 \, \mathrm{mol \, dm^{-3}}$ and $t = 1200 \, \mathrm{s}$.Substitute the values:$[A] = 0.06 \times e^{-(8.15 \times 10^{-4})(1200)}$$[A] = 0.06 \times e^{-0.978}$$e^{-0.978} \approx 0.376$$[A] = 0.06 \times 0.376$$[A] \approx 0.0226 \, \mathrm{mol \, dm^{-3}}$Rounding off, $[A] \approx 0.023 \, \mathrm{mol \, dm^{-3}}$.Answer:The concentration remaining after $1200 \, \mathrm{s}$ is approximately $0.023 \, \mathrm{mol \, dm^{-3}}$, which matches Option A.