A) 1/6 of initial concentration
B) 1/64 of initial concentration
C) 1/12 of initial concentration
D) 1/32 of initial concentration
Correct Answer: B
Solution :
\[{{t}_{1/2}}=10;\] Number of half-life \[=\frac{60}{10}=6=\] half-life \[{{C}_{t}}=\frac{{{C}_{0}}}{{{\left( 2 \right)}^{n}}}=\frac{{{C}_{0}}}{{{\left( 2 \right)}^{6}}}=\left( \frac{{{C}_{0}}}{64} \right)\]
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