Answer
1.5 mg
Work Step by Step
Amount of radionuclide at the beginning: $A_{0}=96\, mg$
Rate constant $k=\frac{0.693}{t_{1/2}}=\frac{0.693}{10.\,min}=0.0693/min$
Time $t=1\,hour=60\,min$
$\ln(\frac{A_{0}}{A})=kt$ where $A$ is the amount of radionuclide remaining.
$\implies \ln(\frac{96\,mg}{A})=0.0693\times60=4.158$
Taking the inverse $\ln$ of both sides, we have
$\frac{96\,mg}{A}=e^{4.158}=63.94$
Or $A= \frac{96\,mg}{63.94}=1.5\,mg$
