Q.Estimate the volume of a water molecule using the data in Example 12.1.

The key idea is to treat the water molecule as occupying a tiny cube whose side length is the average intermolecular separation, then compute its volume.

From Example 12.1, the density of water is $\rho = 1000\ \text{kg/m}^3$ and the molar mass is $M = 18\ \text{g/mol} = 0.018\ \text{kg/mol}$. The number of molecules per unit volume is

$n = \frac{\rho N_A}{M} = \frac{1000 \times 6.022 \times 10^{23}}{0.018} \approx 3.34 \times 10^{28}\ \text{m}^{-3}.$ …

The idea is simple: if you know how much space one mole of water occupies, and you know how many molecules are in that mole, then dividing the molar volume by Avogadro’s number gives the volume per molecule. This is called the molecular volume fraction approach — it treats the molecules as tiny cubes or spheres that pack together to fill the bulk liquid.

From Example 12.1 (which I assume gives the density of water as $1000 \, \text{kg/m}^3$ and the molar mass as $18 \, \text{g/mol}$), we can proceed step by step.

1. Find the volume of one mole of water.

   Density $\rho = 1000 \, \text{kg/m}^3 = 10^6 \, \text{g/m}^3$ (since $1 \, \text{kg} = 1000 \, \text{g}$).

   Molar mass $M = 18 \, \text{g/mol}$.

   Molar volume $V_m = \frac{M}{\rho} = \frac{18 \, \text{g/mol}}{10^6 \, \text{g/m}^3} = 1.8 \times 10^{-5} \, \text{m}^3/\text{mol}$.

2. Divide by Avogadro’s number.

   Avogadro’s number $N_A = 6.022 \times 10^{23} \, \text{mol}^{-1}$.

   Volume per molecule $v = \frac{V_m}{N_A} = \frac{1.8 \times 10^{-5}}{6.022 \times 10^{23}} \, \text{m}^3$.

3. Calculate.

   $v = \frac{1.8}{6.022} \times 10^{-28} \approx 0.299 \times 10^{-28} = 3.0 \times 10^{-29} \, \text{m}^3$.

A quick sanity check on $2.99\times10^{-29}\ \text{m}^3$: treat the molecule as a small sphere and back out its effective radius, $r = \left(\frac{3V}{4\pi}\right)^{1/3} \approx 1.9\times10^{-10}\ \text{m} = 1.9\ \text{\AA}$. This matches the known scale of a water molecule (covalent O–H radii, roughly 1–2 \AA), confirming the density-based estimate is physically sensible. Converting a computed volum …