A) 2
B) 3
C) 4
D) 5
Correct Answer: C
Solution :
[c] \[{{t}_{1/2}}=\frac{0.0693}{{{k}_{1}}}\] Also, \[{{t}_{93.75}}=\frac{2.303}{{{k}_{1}}}\log \frac{100}{100-93.75}\] \[=\frac{2.303}{{{k}_{1}}}\log \frac{100}{6.25}=\frac{4\times 2.303\times \log 2}{{{k}_{1}}}=\frac{4\times 0.693}{{{k}_{1}}}=4{{t}_{1/2}}\]
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