Calculate the mass density of water vapor at $100^oC$ and $1.00\; atm$.
$\rho=\frac{m}{V}$ $\rho$ is an intensive physical parameter, which does not depend on the quantity of the sample!
Consider one mole of water vapor. i.e. $18 \;g$ : $V$ $=$ $\frac{nRT}{P}$ $=$ $\frac{1\cdot 0.082\cdot 373.15}{1}\;L$ $\rho$ $=$ $\frac{m}{V}$ $=$ $\frac{18\cdot 1}{1\cdot 0.082\cdot 373.15}$ $=$ $0.589\frac{g}{L}$
