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?
The molecule length value you are using is the problem. It is more like 300 pm, which is the van der Vaals diameter and includes regions where electron density is significant. You have calculated an internuclear distance, but the nuclei of one molecule won't come close to those of another molecule because of electron-electron repulsion.

This will change your answer by a factor of 8.

See Finney, J. L. The water molecule and its interactions: the interaction between theory, modelling, and experiment. J. Mol. Liq. 2001, 90 (1-3), 303–312. DOI: 10.1016/S0167-7322(01)00134-9.