Q44 / 150 · MHT CET 2026 (19 April Shift 2) · Chemistry
Half life of a first order reaction is $900$ second. If initial concentration of reactant is $0.08\,\text{mol dm}^{-3}$ find concentration that remains after $35$ minute ?
- A$0.159\,\text{mol dm}^{-3}$
- Correct answerB$0.0159\,\text{mol dm}^{-3}$
- C$1.05\,\text{mol dm}^{-3}$
- D$0.759\,\text{mol dm}^{-3}$
Solution
Given $t_{1/2} = 900$ s and $t = 35$ min $= 35 \times 60 = 2100$ s.
Number of half-lives $n = \dfrac{t}{t_{1/2}} = \dfrac{2100}{900} = \dfrac{7}{3}$.
For a first order reaction, the concentration remaining after $n$ half-lives is given by:
$[A] = \dfrac{[A]_0}{2^n}$
Substituting the given values:
$[A] = \dfrac{0.08}{2^{7/3}}$
$[A] = \dfrac{0.08}{4 \times 2^{1/3}}$
Since $2^{1/3} \approx 1.26$:
$[A] \approx \dfrac{0.08}{4 \times 1.26} = \dfrac{0.08}{5.04} \approx 0.0159 \text{ mol dm}^{-3}$
Answer: $0.0159\,\text{mol dm}^{-3}$
Number of half-lives $n = \dfrac{t}{t_{1/2}} = \dfrac{2100}{900} = \dfrac{7}{3}$.
For a first order reaction, the concentration remaining after $n$ half-lives is given by:
$[A] = \dfrac{[A]_0}{2^n}$
Substituting the given values:
$[A] = \dfrac{0.08}{2^{7/3}}$
$[A] = \dfrac{0.08}{4 \times 2^{1/3}}$
Since $2^{1/3} \approx 1.26$:
$[A] \approx \dfrac{0.08}{4 \times 1.26} = \dfrac{0.08}{5.04} \approx 0.0159 \text{ mol dm}^{-3}$
Answer: $0.0159\,\text{mol dm}^{-3}$
Hence, the correct answer is option: 'B'.