Calculate the mass density of water in gaseous phase at $100^oC$ and $1.00$ atm.
$\rho=\frac{m}{V}$
$\rho$ is an intensive physical parameter, which does not depend on how much of the sample one takes!
Let's take one mole of water vapour, i.e. $18$ g
$V$ $=$ $\frac{nRT}{P}$ $=$ $\frac{1\cdot 0.082\cdot 373.15}{1}$ L
$\rho$ $=$ $\frac{m}{V}$ $=$ $\frac{18\cdot 1}{1\cdot 0.082\cdot 373.15}$ $=$ $0.589\frac{g}{L}$
