A) \[8.0\,\,g\]
B) \[4.0\,\,g\]
C) \[12.0\,\,g\]
D) \[16.0\,\,g\]
Correct Answer: B
Solution :
\[\because \,\,{{t}_{1/2}}=3\,\,h\], \[{{N}_{0}}=256\,\,g\] and \[t=18\,\,h\] \[\because \] \[{{t}_{1/2}}=\frac{0.693}{\lambda }\] \[\Rightarrow \] \[3=\frac{0.693}{\lambda }\] \[\therefore \] \[\lambda =0.231\,\,{{h}^{-1}}\] On applying the equation \[\lambda =\frac{2.303}{t}\log \frac{{{N}_{0}}}{N}\] \[\Rightarrow \] \[0.231=\frac{2.303}{18}\log \frac{256}{N}\] \[\Rightarrow \] \[\frac{0.231\times 18}{2.303}=\log \frac{256}{N}\] \[\Rightarrow \] \[1.8054=\log \frac{256}{N}\] \[\Rightarrow \] \[63.8956=\frac{256}{N}\] \[\Rightarrow \] \[N=\frac{256}{63.8956}\] \[\therefore \] \[N=4.00\](approximately)
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