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The half-life of Iodine- $131$ is approximately $8$ days. What is the amount of Iodine- $131$ left from a $35$ gram sample after $32$ days?

Answer
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Hint: We must know that the atomic half-life period of a radioactive isotope discloses to you how long should pass all together for half of the iotas present in an underlying example to go through radioactive decay. Generally, the half-life advises you at what time stretches you can expect an underlying example of a radioactive isotope to be halved.

Complete step by step answer:
Given,
The mass of Iodine- $131$ is given as $35g.$
The Half-life of Iodine- $131$ is $8d.$
The time interval is $32d.$
The number of half-life can be calculated by dividing the time interval by the half-life of the isotope.
The number of half-life can be calculated as follows,
$32d \times \dfrac{{1half - life}}{{8d}} = 4$ half-life
By multiplying half-lives with the initial mass one can calculate the mass of substance remains after half-lives,
The mass of substance can be calculated as follows,
The amount of substance remains after four half-lives (m)${t{ = 35g \times }}\dfrac{{{1}}}{{{2}}} \times \dfrac{{{1}}}{{{2}}} \times \dfrac{{{1}}}{{{2}}} \times \dfrac{{{1}}}{{{2}}}$
$m = 2.2g$

The mass of Iodine-$131$ remains after ${{32d}}$ is $2.2g$

Note: One can define the activity as the process in which number of disintegrations per second or the number of unstable atomic nuclei that decay per second in a given sample. One can also find out the activity from the half-life period.
Example:
Let us assume the activity of the sample is ${{240}}\,{{mci}}{{.}}$
The time interval is $32\,{{days}}{{.}}$
The half-life of iodine is ${{8}}{{.0}}\,{{days}}{{.}}$
The number of half-life can be calculated by dividing the time interval by the half-life of the isotope.
The number of half-life can be calculated as follows,
$32d \times \dfrac{{1half - life}}{{8d}} = 4$ half-life
By multiplying half-lives with the initial activity one can calculate the activity of the sample remains after half-lives.
The mass of substance can be calculated as follows,
The amount of substance remains after four half-lives (m)${{ = 240mci \times }}\dfrac{{{1}}}{{{2}}} \times \dfrac{{{1}}}{{{2}}} \times \dfrac{{{1}}}{{{2}}} \times \dfrac{{{1}}}{{{2}}}$
The activity of the sample after $32days$ is $15mci$