A) \[8.0\text{ }g\]
B) \[4.0\text{ }g\]
C) \[12.0\text{ }g\]
D) \[16.0\text{ }g\]
Correct Answer: B
Solution :
\[\because \] \[{{t}_{1/2}}=3h,\] \[{{N}_{0}}=256g\] and \[t=18h\] \[\because \] \[{{t}_{1/2}}=\frac{0.693}{\lambda }\] \[\Rightarrow \] \[3=\frac{0.693}{\lambda }\] \[\therefore \] \[\lambda =0.231{{h}^{-1}}\] 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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