A) 399 min
B) 410 min
C) 250 min
D) 120 min
Correct Answer: A
Solution :
First calculate the values of k \[K=\frac{0.693}{{{T}_{1/2}}}=\frac{0.693}{120}=5.77\times {{10}^{-3}}\,{{\min }^{-1}}\] Now we know that for a first-order reaction \[K=\frac{2.303}{t}\,\log \,\left( \frac{a}{a-x} \right)\] Here the initial concentration \[\operatorname{a} =100\] and \[n =90\] \[\therefore \,\,\,\,\,5.77\times {{10}^{-3}}=\frac{2.303}{t}\,\log \,\frac{100}{100-90}\] \[t=\frac{2.303}{5.77\times {{10}^{-3}}}\,\,\log \,\frac{{{10}^{-3}}\times {{10}^{3}}}{6.023\times {{10}^{23}}\times \frac{1}{3600}\,hr}\] \[\operatorname{Solving} t = 399 minute\]
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