A flask is filled with equal moles of A and B. The half lives of A and B are 100 s and 50 s respectively and are independent of the initial concentration. The time required for the concentration of A to be four times that of B is ___________ s.

(Given : ln 2 = 0.693)

$\mathrm{k}_{\mathrm{A}}=\frac{\ln 2}{100} ; \mathrm{k}_{\mathrm{B}}=\frac{\ln 2}{50}$<br/><br/> $\mathrm{A}_{\mathrm{t}}=\mathrm{A}_0 \times \mathrm{e}^{-\mathrm{k}_{\mathrm{A}} \mathrm{t}}$<br/><br/> $\mathrm{A}_{\mathrm{t}}=\mathrm{A}_0 \times \mathrm{e}^{\left(\frac{-\ln 2}{100} \times \mathrm{t}\right)}$<br/><br/> $\mathrm{B}_{\mathrm{t}}=\mathrm{B}_0 \times \mathrm{e}^{\left(\frac{-\ln 2}{50} \times \mathrm{t}\right)}$<br/><br/> $\mathrm{A}_0=\mathrm{B}_0$<br/><br/> $\& \mathrm{~A}_{\mathrm{t}}=4 \mathrm{~B}_{\mathrm{t}}$<br/><br/> $\mathrm{e}^{-\frac{\ln 2}{100} \times \mathrm{t}}=4 \times \mathrm{e}^{-\frac{\ln 2}{50} \times \mathrm{t}}$<br/><br/> $\mathrm{e}^{\frac{\ln 2}{100} \times \mathrm{t}}=4$<br/><br/> $\mathrm{e}^{\frac{\ln 2}{100} \times \mathrm{t}}=4$<br/><br/> $\frac{\ln 2}{100} \times \mathrm{t}=\ln 4=2 \ln 2$<br/><br/> $\mathrm{t}=200 ~ \mathrm{sec}$