A) 0.5 mg
B) 0.726 mg
C) 0.126 mg
D) 0.236 mg
Correct Answer: D
Solution :
Given, half-life period, \[{{t}_{1/2}}=4.8\,\,\min \] initial amount, \[{{N}_{0}}=1\,mg\] total time, \[t=10\text{ }min\] \[\Rightarrow \] \[k=\frac{0.693}{{{t}_{1/2}}}=\frac{0.693}{4.8}\] \[\Rightarrow \] \[k=\frac{2.303}{t}\log \frac{{{N}_{0}}}{N}\] or \[\frac{0.693}{4.8}=\frac{2.303}{10}\log \frac{1}{N}\] \[\log \frac{1}{N}=0.62689\] \[N=0.236\,mg\]
You need to login to perform this action.
You will be redirected in
3 sec
