At 25^{o}C (current temperature used in the rest of our pH study), the voluminal mass of water is worth:
$\rho_{H_2O}$ $=$ $0.99704 \frac{g}{cm^3}$

As the mass of one liter of water (= 1 dm^{3}) is $997.04\; g$, we can calculate the molarity of water:
Number of moles of water in $1\; L$ of water
= $\frac{m}{M}$
= $\frac{997.04}{18.01}$
= $\frac{997.04}{18.01}$
= $55.36 $ mol

At 25^{o}C
$c_{H_2O}^*$ $=$ $55.36 \frac{mol}{L}$

$^*$ We will define later the difference between: $c_ {H_2O}$ (formal molarity) and $[H_2O]$ (actual molarity)