A) 1.023M
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, \[\log \frac{[{{A}_{0}}]}{A}=\frac{kxt}{2.303}=\frac{0.693}{3}\times \frac{8}{2.303}=0.804\] \[\frac{{{[A]}_{0}}}{[A]}=\text{Antilog}\,\text{0}\text{.8024}\,\text{=}\,6.345\] \[{{[A]}_{0}}=1\,M;\] \[[A]=\frac{1}{6.345}=0.1576\,M\]
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