Question

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. \(0.159\,\text{mol dm}^{-3}\)
• B. \(0.0159\,\text{mol dm}^{-3}\)
• C. \(1.05\,\text{mol dm}^{-3}\)
• D. \(0.759\,\text{mol dm}^{-3}\)

Hint:

Convert 35 minutes to seconds, count the half lives, then divide the initial concentration by 2 to that power.

Answer 1

The Correct Option is B

Step 1: Use half lives
Time elapsed in half lives: $2100/900 = 7/3$.

Step 2: Halve step by step
After 2 half lives (1800 s): $0.08/4 = 0.02$. A third of another half life remains, which reduces by the factor $2^{-1/3} = 0.794$.

Step 3: Compute
$0.02 \times 0.794 = 0.01587 \approx 0.0159$ mol dm$^{-3}$.

Step 4: Compare
Options (C) and (D) exceed the starting 0.08, which is impossible, and option (A) is also above 0.08. So only 0.0159 fits.

Final Answer:
Concentration left is 0.0159 mol per dm^3. This is option (B). \[ \boxed{\text{(B) }0.0159\ \text{mol dm}^{-3}} \]