A) 693.0 second
B) 238.6 second
C) 138.6 second
D) 346.5 second
Correct Answer: C
Solution :
| [c] \[{{t}_{1/2}}=\frac{0.693}{{{10}^{-2}}}\]second |
| For the reduction of 20 g of reactant to 5 g, two \[{{t}_{1/2}}\]is required. |
| \[\therefore \] \[t=2\times \frac{0.693}{{{10}^{-2}}}\,\text{seecond}\] |
| \[=138.6\,\text{second}\] |
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