A) \[30\text{ }min\]
B) \[40min\]
C) \[38min\]
D) \[45min\]
Correct Answer: B
Solution :
According to the 1st order of kinetics, \[t=\frac{2.303}{\lambda }\log \frac{a}{a-x}\] \[\frac{{{t}_{19%}}}{{{t}_{10%}}}=\frac{\log \frac{100}{100-19}}{\log \left( \frac{100}{100-10} \right)}\] \[=\frac{\log 100-\log 81}{\log \frac{10}{9}}\] \[\frac{{{t}_{19%}}}{20}=\frac{\log {{10}^{2}}-\log {{3}^{4}}}{\log 10-\log {{3}^{2}}}\] \[{{t}_{19%}}=\frac{(2-4\times 0.477)\times 20}{1-2\times 0.477}\] \[=40\min \]
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