A) 48 g
B) 12 g
C) 24 g
D) 6g
Correct Answer: A
Solution :
Key Idea \[\frac{N}{{{N}_{0}}}={{\left( \frac{1}{2} \right)}^{n}}\] where, n = number of half-life period \[{{N}_{0}}\] = initial corcentration N = concentration left after n half-life period and \[T=n\propto {{t}_{1/2}}\] where, \[{{t}_{1/2}}=\] half-life period T = total time Here, \[{{t}_{1/2}}=3\] days, T = 12 days \[\therefore \] \[n=\frac{12}{3}=4\] \[N=3\,g\] \[\therefore \] \[\frac{3}{{{N}_{0}}}={{\left( \frac{1}{2} \right)}^{4}}\] \[\frac{3}{{{N}_{0}}}=\frac{1}{16}\] \[{{N}_{0}}=48\,\,g\]
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