A) 25 g
B) 50 g
C) \[12.5\]g
D) \[6.25\]g
Correct Answer: C
Solution :
\[C={{C}_{0}}{{\left( \frac{1}{2} \right)}^{\gamma }}\] \[y=\frac{total\,time}{half-life}=3={{\left( \frac{1}{2} \right)}^{3}}=12.5\,\,g\] For first order chemical kinetics \[K=\frac{2.303}{t}\log \frac{[{{R}_{0}}]}{[R]}\] where, \[[{{R}_{0}}]\] is initial concentration, [R] is concentration of reactant after t time. Relation between half-life of a substance with K is \[K=\frac{0.69}{{{t}^{1/2}}}=\frac{0.69}{5}\] On placing the value of k in above equation \[\frac{0.69}{5}=\frac{2.303}{15}\log \frac{100}{[A]}\] \[\log \frac{100}{[A]}=\frac{0.69\times 15}{5\times 2.303}=0.898\] \[\log 100-\log [A]=0.898\] \[2-0.898=\log [A]\] \[\log [A]=1.1\] \[[A]=\] antilog 1.1 \[[A]=12.5g\]
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