Answer: 16.81 min
Step-by-step explanation:
Given
Initial concentration [tex]A_o=970\ gm[/tex]
Half-time [tex]T_{\frac{1}{2}}=6\ min[/tex]
Decay at any time t is given by
[tex]A=A_o2^{\dfrac{-t}{T_{\frac{1}{2}}}}[/tex]
put values
[tex]\Rightarrow 139=970\cdot 2^{\dfrac{-t}{6}}\\\\\Rightarrow 2^{\dfrac{t}{6}}=6.978\\[/tex]
Taking log both sides
[tex]\Rightarrow \dfrac{t}{6}\times \ln (2)=\ln (6.978)\\\\\Rightarrow t=6\times 2.802\\\Rightarrow t=16.81\ \text{min}[/tex]