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Q.
The half-life of a first order reaction is $ 10$ minutes. If initial amount is $0.08 \,mol/litre$ and concentration at some instant is $0.01\, mol/litre$, then t is
Chemical Kinetics
Solution:
$k=\frac{0.693}{t_{1/2}}=\frac{0.693}{10}$
$t =\frac{2.303}{k} log \frac{0.08}{0.01}=\frac{2.303}{0.693}log 8=30 \,minutes$
Alternatively,
Amount left after $n$ half-lives,$0.01 =\frac{0.08}{2^{n}}$
$2^{n} =\frac{0.08}{0.01}=8$
Then, total time = $3 \times10 = 30\, minutes $