# Calculating the volume of 1 mole of liquid water

I am doing some rough calculations to test my understanding of elementary chemistry. In what follows, I calculate the volume of one mole of liquid water at about $\pu{4 ^\circ C}$ and $\pu{1 atm}$:

Method 1: Using the molar mass and density of water

The molar mass of water is $\pu{18 \times 10^-3 kg mol-1}$, and the density is $\pu{1 \times 10^3 kg m-3}$. Hence, the volume should be

$$V = \frac{(\pu{1 mol})(\pu{18 \times 10^-3 kg mol-1})}{\pu{1 \times 10^3 kg m-3}} = \pu{1.8 \times 10^-5 m3}.$$

Method 2: Using the size of an individual water molecule

If we model the mole of water as a cube, since there are $6.022 \times 10^{23}$ water molecules, each side of the cube would have $8.44 \times 10^{7}$ molecules.

Using the water data page on Wikipedia, I used trigonometry to calculate the H–H distance as being approximately $\pu{150 \times 10^-12 m}$. I would think the dimensions do not change too much when there are many molecules.

From this, I find that the volume is

$$V = [(8.44 \times 10^7) \times (\pu{150 \times 10^-12 m})]^3 = \pu{2.05 \times 10^-6 m3}$$

which is an order of magnitude smaller than Method 1.

What is the reason for this discrepancy?