The density of water is $1000 \; kg \; m^{-3}$. The density of water vapour at $100^{\circ} C$ and $1 \; atm$ pressure is $0.6 \; kg \; m^{-3}$. The volume of a molecule multiplied by the total number gives what is called molecular volume. Estimate the volume of a water molecule.

(N/A) In the liquid phase,the molecules of water are quite closely packed. The density of a water molecule may therefore be regarded as roughly equal to the density of bulk water $= 1000 \; kg \; m^{-3}$.
To estimate the volume of a water molecule,we need to know the mass of a single water molecule.
We know that $1 \; mole$ of water has a mass approximately equal to $(2 + 16) \; g = 18 \; g = 0.018 \; kg$.
Since $1 \; mole$ contains about $6 \times 10^{23}$ molecules (Avogadro's number),the mass of a single molecule of water is $(0.018) / (6 \times 10^{23}) \; kg = 3 \times 10^{-26} \; kg$.
Therefore,a rough estimate of the volume of a water molecule is:
Volume of a water molecule $= (3 \times 10^{-26} \; kg) / (1000 \; kg \; m^{-3}) = 3 \times 10^{-29} \; m^3$.
Assuming the molecule is spherical,$V = (4/3) \pi r^3$,which gives a radius $r \approx 2 \times 10^{-10} \; m = 2 \; \mathring{A}$.