A) \[1.023\,\,M\]
B) \[0.8725\,\,M\]
C) \[0.023\,\,M\]
D) \[0.1576\,\,M\]
Correct Answer: D
Solution :
For first order reactions, \[k=\frac{0.693}{{{t}_{1/2}}}=\frac{0.693}{(3.0)}\] \[t=\frac{2.303}{K}\log \frac{[{{A}_{0}}]}{[A]}\] Thus,\[\log \frac{[{{A}_{0}}]}{[A]}=\frac{kxt}{2.303}=\frac{0.693}{3}\times \frac{8}{2.303}=0.804\] \[\frac{{{[A]}_{0}}}{[A]}=\text{Antilog}\,\,0.8024=6.345\] \[{{[A]}_{0}}=1\,\,M;\] \[[A]=\frac{1}{6.345}=0.1576\,\,M\]
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