A) \[0.0075\text{ }M\]
B) \[0.06\text{ }M\]
C) \[0.03\text{ }M\]
D) \[0.015\text{ }M\]
Correct Answer: A
Solution :
\[{{t}_{1/2}}=4s\,\,\,T=16s\] \[n=\frac{T}{{{t}_{1/2}}}=\frac{16}{4}=4\] \[(\therefore \,\,T=n\times {{t}_{{}^{1}/{}_{2}}})\] \[[A]={{[A]}_{o}}{{\left( \frac{1}{2} \right)}^{n}}=0.12\times {{\left( \frac{1}{2} \right)}^{4}}=\frac{0.12}{16}=0.0075\,M\] Where \[{{[A]}_{o}}\] initial concentration and \[[A]\] = concentration left after time t
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