Calculate the volume of vapour of $1\,g$ of water at $1\,atm$ pressure.

To find the volume of $1\,g$ of water vapour at $1\,atm$ pressure,we use the ideal gas law: $PV = nRT$.
Here,$P = 1\,atm = 1.013 \times 10^5\,Pa$.
The mass of water is $m = 1\,g = 10^{-3}\,kg$.
The molar mass of water $(H_2O)$ is $M = 18\,g/mol = 18 \times 10^{-3}\,kg/mol$.
The number of moles $n = \frac{m}{M} = \frac{1}{18}\,mol$.
The universal gas constant $R = 8.314\,J/(mol \cdot K)$.
Assuming the temperature $T = 373.15\,K$ (boiling point of water at $1\,atm$):
$V = \frac{nRT}{P} = \frac{(1/18) \times 8.314 \times 373.15}{1.013 \times 10^5}$.
$V \approx 1.70 \times 10^{-3}\,m^3 = 1.70\,litres$.