Q.What is the average distance between atoms (interatomic distance) in water? Use the data given in Examples 12.1 and 12.2.

Concept: Use the volume-expansion result from Example 12.1 together with the molecular-volume estimate from Example 12.2 — spacing scales as the cube root of the volume available per molecule.

1. Liquid-phase spacing (Example 12.2): $V_{\text{molecule}}\approx3.0\times10^{-29}\ \text{m}^3$, so $d_{\text{liquid}}=(V_{\text{molecule}})^{1/3}\approx4\times10^{-10}$ m.
2. Volume increase on vaporisation (Example 12.1): vapour occupies $1000/0.6\approx1.67\times10^3$ times the liquid volume for the same mass, so the volume available per molecule increases by that same factor. …

The estimation method

We want the typical centre-to-centre distance between neighbouring molecules in liquid water. Since water molecules are essentially close-packed in the liquid, a standard estimate treats each molecule as sitting at the centre of its own small cube of side $d$, with $d^3$ equal to the volume it effectively occupies in the bulk liquid.

Step-by-step calculation

1. Data (from Examples 12.1 and 12.2 -- standard NCERT figures for water):

• Density: $\rho = 1000\ \text{kg/m}^3$
• Molar mass: $M = 18\ \text{g/mol} = 18\times10^{-3}\ \text{kg/mol}$
• Avogadro's number: $N_A = 6.02\times10^{23}\ \text{mol}^{-1}$

2. Volume of one mole of water.

3. Volume occupied by one molecule.

4. Take the cube root to get the spacing.

5. Express in convenient units. …

A faster route skips computing the vapour-phase volume directly: reuse the liquid-phase spacing from Example 12.2 ($d_{liquid}\approx4\times10^{-10}\ \text{m}$) and just scale it by the cube root of the volume-expansion factor from Example 12.1. Since $V_{vapour}/V_{liquid}\approx1.67\times10^3$ and spacing scales as $V^{1/3}$, the vapour spacing is $d_{vapour}=d_{liquid}\times(1.67\times10^3)^{1/3}\approx d_{liquid}\times10 \approx4\times10^{-9}\ \text{m}=40\ \text{\AA}$. The general shortcut worth remembering — 'a vol …