Answer
$M=71.1 \frac{g}{mol}$, which corresponds to the molecular chlorine, $Cl_{2}$.
Work Step by Step
$m=0.48g$
$V=367ml=0.367dm^{3}$
$p=365torr=365\times 133.3Pa = 48654.5Pa=48.6545kPa$
$T=45^{\circ}C=(45+273.15)K=318.15K$
---$pV=nRT \implies pV=\frac{m}{M}RT \implies M=\frac{mRT}{pV}=\frac{0.48g\times 8.314\frac{J}{K\times mol}\times 318.15K}{48.6545kPa\times 0.367dm^{3}}=71.1\frac{g}{mol}$
This corresponds to the molecular mass of chlorine, $Cl_{2}$. Since there are not many gaseous elements, this is the only reasonable option.
