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If 168 grams of radioactive iron (59-Fe) is recived in the lab today, what percentage of...

If 168 grams of radioactive iron (59-Fe) is recived in the lab today, what percentage of the original is left after 135 day?
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Answer #1

half life of Fe -59 is 45 days

rate constant = 0.693 / half life

                    = 0.693 / 45

                    = 0.0154 day-1

k = 1/t * ln (A /At)

0.0154 = 1/135 * ln (168 / At)

At = 21

remaining mass = 21 g = left

percent of original is left = 21 x 100 / 168

                                       = 12.5 %

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