## Chemistry: A Molecular Approach (3rd Edition)

$1.328125\times 10^{-18}g$
We know that $\rho=\frac{m}{V}$ Where $\rho$, $m$ and $V$ are density, mass and volume of oxygen respectively The above equation can be rearranged as $m =V\times \rho$ We plug in the known values to obtain: $m=1.5625\times 10^{-20}\times 85=1.328125\times 10^{-18}g$