Question

Open in App

Updated on : 2022-09-05

Solution

Verified by Toppr

$k=28.10.693 =0.0247/year$

(i) After 10 years, let $xμ$ be the concentration of $_{90}Sr$.

$k=t2.303 log[A][A]_{0} $

$0.0247=10years2.303 log(x1 )$

$x=0.782μg$

(ii) After 60 years, let $yμ$ be the concentration of $_{90}Sr$.

$k=t2.303 log[A][A]_{0} $

$0.0247=60years2.303 log(y1 )$

$y=0.228μg$

Video Explanation

Solve any question of Chemical Kinetics with:-

Was this answer helpful?

0

0