Answer:
[tex]\large \boxed{\text{4.0 g}}[/tex]
Explanation:
The half-life of chromium-51 (28 da) is the time it takes for half of it to decay.
After one half-life, half of the original amount will remain.
After a second half-life, half of that amount will remain, and so on.
We can construct a table as follows:
[tex]\begin{array}{ccccc}\textbf{No. of} & &\textbf{Fraction} &\textbf{Mass}\\ \textbf{Half-lives} & \textbf{t/da} &\textbf{Remaining}&\textbf{Remaining/g}\\0 & 0 & 1 &32.0\\\\1 & 28 & \dfrac{1}{2} & 16.0\\\\2 & 56 & \dfrac{1}{4} & 8.00\\\\3 & 84 & \dfrac{1}{8} & 4.00\\\\4 & 112 & \dfrac{1}{16} & 2.00\\\\\end{array}[/tex]
We see that, after 84 da (three half-lives), ⅛ of the original mass remains.
[tex]\text{Mass remaining} = \left(\dfrac{1}{2}\right)^{3} \times \text{32 g}= \dfrac{1}{8}\times\text{32.0 g} = \large \boxed{\textbf{4.00 g}}[/tex]
