**Question:**

A flask contains a mixture of compounds $\mathrm{A}$ and

B. Both compounds decompose by first-order kinetics. The half-lives for $\mathrm{A}$ and $\mathrm{B}$ are $300 \mathrm{~s}$ and $180 \mathrm{~s}$, respectively. If the concentrations of $\mathrm{A}$ and $\mathrm{B}$ are equal initially, the time required for the concentration of $A$ to be four times that of $B($ in $s):($ Use $\ln 2=0.693)$

Correct Option: , 4

**Solution:**

$[\mathrm{A}]_{\mathrm{t}}=4[\mathrm{~B}]_{\mathrm{t}}$

$[\mathrm{A}]_{0} \mathrm{e}^{-\left(\ln ^{2} / 300\right)^{t}}=4[\mathrm{~B}]_{0} \mathrm{e}^{(-\ln 2 / 180) t}$

$e^{\left(\frac{\ln ^{2}}{180}-\frac{\ln ^{2}}{300}\right)}=4$

$\left(\frac{\ln ^{2}}{180}-\frac{\ln ^{2}}{300}\right) \mathrm{t}=\ln 4$

$\left(\frac{1}{180}-\frac{1}{300}\right) \mathrm{t}=2 \Rightarrow \mathrm{t}=\frac{2 \times 180 \times 300}{120}=900 \mathrm{sec}$