A) zero
B) 25
C) 12.5
D) 6.25
Correct Answer: B
Solution :
We know that \[Cn\frac{C{}^\circ }{{{2}^{n}}}\] Here, \[n=\]number of half lives = 2 \[Cn=\]concentration after \[n\]halves \[{{C}^{0}}=\]initial concentration \[=1\] \[\therefore \] \[{{C}_{n}}=\frac{1}{{{2}^{2}}}=\frac{1}{4}\] \[=0.25=25%\]
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