Answer
$\approx 3.2$ hours
Work Step by Step
A decay model has the form $Q(t)=Q_{0}e^{-kt}$.
The decay constant k and half-life $t_{h}$ are related by
$t_{h}k=\ln 2$
Substitute $t_{h}=2$ and solve for k.
$2k=\ln 2$
$k=\displaystyle \frac{\ln 2}{2}\approx 0.3466$
So,
$Q(t)\approx Q_{0}e^{(-0.3466)t}$
Now, $Q_{0}=300$ g, $Q(t)=100$ g. Substituting, we solve for t.
$100=300 e^{(-0.3466)t} \qquad/\div 300$
$e^{(-0.3466)t}=\displaystyle \frac{1}{3}\qquad/\ln(..)$
$-0.3466t=\displaystyle \ln(\frac{1}{3})$
$ t=\displaystyle \frac{\ln(\frac{1}{3})}{-0.3466}\qquad$ ... round to nearest tenth.
$\approx 3.2$ hours
