A) \[50\,\,\min \]
B) \[100\,\,\min \]
C) \[99\,\,\min \]
D) \[150\,\,\min \]
Correct Answer: C
Solution :
Key Idea: \[k=\frac{2.303}{t}\log \frac{a}{a-x}\] Given,\[t=15,\,\,a=100,\,\,a-x=50\] \[\therefore \] \[k=\frac{2.303}{15}\log \frac{100}{50}\] \[k=\frac{2.303}{15}\log 2\] ? (i) Also\[t=?,\,\,a=100,\,\,a-x=100-99=1\] \[\therefore \] \[k=\frac{2.303}{t}\log \frac{100}{1}\] or \[k=\frac{2.303}{t}\log 100\] ... (ii) From Eqs. (i) and (ii). \[\frac{2.303}{15}\log 2=\frac{2.303}{t}\log 100\] or \[\frac{1}{15}\log 2=\frac{1}{t}\log 100\] or \[\frac{0.3010}{15}=\frac{2}{t}\] or \[t=\frac{15\times 2}{0.3010}\] \[=99\min \]
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