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# The molar concentration (molarity) of water

At 25oC (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 dm3) 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 25oC $c_{H_2O}^*$ $=$ $55.36 \frac{mol}{L}$

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