Given:
Half life = 12.3 years
use relation between rate constant and half life of 1st order
reaction
k = (ln 2) / k
= 0.693/(half life)
= 0.693/(12.3)
= 5.634*10^-2 years-1
Given:
[A]o = 100 (Let initial amount be 100)
t = 1.32 years
k = 5.634*10^-2 years-1
use integrated rate law for 1st order reaction
ln[A] = ln[A]o - k*t
ln[A] = ln(1*10^2) - 5.634*10^-2*1.32
ln[A] = 4.605 - 5.634*10^-2*1.32
ln[A] = 4.531
[A] = e^(4.531)
[A] = 92.8
92.8 of 100 remains.
This is 92.8 %
Answer: 93 %
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