Answer
$13.8\,\mu g$
Work Step by Step
Quantity that should remain: $N= 10.0\,\mu g$
Decay constant $k=\frac{0.693}{t_{1/2}}=\frac{0.693}{4.5\times24\,h}$
Time $t= 50\,h$
$\ln(\frac{N_{0}}{N})=kt$ where $N_{0}$ is the quantity that the chemist must order.
$\implies \ln(\frac{N_{0}}{10.0\,\mu g})=(\frac{0.693}{4.5\times24\,h})\times50\,h=0.3208$
Taking the inverse $\ln$ of both sides, we have
$\frac{N_{0}}{10.0\,\mu g}=e^{0.3208}=1.378$
Or $N_{0}=1.378\times10.0\,\mu g=13.8\,\mu g$
