Question

years. How long wil take for 30% of the C-14 atons in a sample of C-14 to decay? My Answers Give Up If a sample of C-14 initially contains 1.1 mmol of C-14, how many millimoles will be left after 2255 years? mmol

Answer 1

A)

Given:

Half life = 5730 yr

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

k = (ln 2) / k

= 0.693/(half life)

= 0.693/(5730)

= 1.209*10^-4 yr-1

we have:

[C-14]o = 100 (Let initial amount be 100)

[C-14] = 70.0 (amount decayed is 30%. So, remaining is 70 %)

k = 1.209*10^-4 yr-1

use integrated rate law for 1st order reaction

ln[C-14] = ln[C-14]o - k*t

ln(70) = ln(100) - 1.209*10^-4*t

4.2485 = 4.6052 - 1.209*10^-4*t

1.209*10^-4*t = 0.3567

t = 2949.1305 yr

t = 2.9*10^3 years

Answer: 2.9*10^3 years

B)

[C-14]o = 1.1 mmol

t = 2255.0 yr

k = 1.209*10^-4 yr-1

use integrated rate law for 1st order reaction

ln[C-14] = ln[C-14]o - k*t

ln[C-14] = ln(1.1) - 1.209*10^-4*2255

ln[C-14] = 0.0953 - 1.209*10^-4*2255

ln[C-14] = -0.1774

[C-14] = 0.8374 mmol

Answer: 0.84 mmol