Answer
a) $9.816\times 10^{11}$ oxygen molecules
b) $5.235\times 10^{-11} g$
Work Step by Step
$V=500ml = 0.5dm^{3}=5\times 10^{-4}m^{3}$
$p=2.5\times 10^{-7}torr=2.5\times 10^{-7}\times 133.3 Pa=3.3325\times 10^{-5} Pa$
$T=1225K$
------------a) $N=?$
b) $m=?$
a) $pV=nRT \implies pV=\frac{N}{N_{A}}RT \implies N=\frac{pVN_{A}}{RT}=\frac{3.3225\times 10^{-5}Pa\times 5\times 10^{-4}m^{3}\times 6\times 10^{23} \frac{1}{mol}}{8.314 \frac{J}{K\times mol}\times 1225K}=9.816\times 10^{11}$
b) $n=\frac{N}{N_{A}}=\frac{9.816\times 10^{11}}{6\times 10^{23} \frac{1}{mol}}=1.636\times 10^{-12} mol$
$m=M(O_{2})\times n=32\frac{g}{mol}\times 1.636\times10^{-12}mol=5.235\times 10^{-11}g$
