Answer
$56.2\,\mu g$
Work Step by Step
We find:
$V=7.12\,\mu L\times\frac{10^{-6}\,L}{1\,\mu L}\times\frac{10^{-3}\,m^{3}}{1\,L}=7.12\times10^{-9}\,m^{3}$
$T=(22+273.15)\,K=295.15\,K$
$P=8.72\times10^{4}\,Pa$
$R=8.314472\,J\cdot mol^{-1}K^{-1}$
$PV=nRT$ (Ideal gas law)
$\implies n=\frac{PV}{RT}=\frac{(8.72\times10^{4}\,Pa)(7.12\times10^{-9}\,m^{3})}{(8.314472\,J\cdot mol^{-1}K^{-1})(295.15\,K)}$
$=2.53\times10^{-7}\,mol$
The mass of radon gas= $2.53\times10^{-7}\,mol\times\frac{222\,g}{1\,mol\,Rn\,gas}$
$=5.62\times10^{-5}\,g=56.2\,\mu g$
