A) 4.68 g
B) 46.8 g
C) 9.375 g
D) 93.75 g
Correct Answer: A
Solution :
Number of half-lives \[n=\frac{t}{T}=\frac{18}{3}=6\] Amount remained after n half-lives \[M={{M}_{0}}{{\left( \frac{1}{2} \right)}^{n}}\] \[Given,{{M}_{0}}=300\,g\] \[\therefore M=300\,{{\left( \frac{1}{2} \right)}^{6}}\] \[=300\times \frac{1}{64}=4.68\,g\] Alternative: Total time of decay given \[t=\frac{2.303}{\lambda }{{\log }_{10}}\left( \frac{300}{M} \right)\] but \[\lambda =\frac{0.693}{T}\] \[=\frac{0.693}{3}=0.231/h\] \[\therefore \] \[t=\frac{2.303}{0.231}{{\log }_{10}}\left( \frac{300}{M} \right)\] Given, \[t=18\,h\] So, \[18=\frac{2.303}{0.231}{{\log }_{10}}\left( \frac{300}{M} \right)\] or \[{{\log }_{10}}\left( \frac{300}{M} \right)=\frac{0.231}{2.303}\times 18\] or \[\frac{300}{M}={{(10)}^{1.8}}\] or \[M=\frac{300}{{{(10)}^{1.8}}}=4.68\,g\]
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