Question

32P is a radioactive isotope with a half-life of 14.3 days. If you currently have 75.7 g of 32P, how much 32P was present 6.00 days ago?

Answer 1

we have:

Half life = 14.3 days

use relation between rate constant and half life of 1st order reaction

k = (ln 2) / k

= 0.693/(half life)

= 0.693/(14.3)

= 4.846*10^-2 days-1

we have:

[P] = 75.7 g

t = 6.0 days

k = 4.846*10^-2 days-1

use integrated rate law for 1st order reaction

ln[P] = ln[P]o - k*t

ln(75.7) = ln[P]o - 4.846*10^-2*6

ln[P]o = 4.3268 + 4.846*10^-2*6

ln[P]o = 4.6175

[P]o = 101 g

Answer: 101 g