# How can I calculate molality of pure water?

Apr 9, 2015

Sorry, but there actually is an answer. Just like there's a real molar concentration for water by itself ($\text{55.348 M}$), there is a real molality for water by itself ($\text{55.510 m}$).

Let's say you had $\text{1 L}$ of water. At ${25}^{\circ} \text{C}$, the density of water is $\text{0.9970749 g/mL}$, so that's $\text{0.9970749 kg}$.

At that amount of water, the number of moles is

$\text{997.0749 g" / ("18.0148 g/mol") = "55.348 mols}$

molality:

$\text{mol water"/"kg water}$ = "55.348 mols"/"0.9970749 kg" = "55.50991407" ~~ color(blue)("55.510 m")

molarity:

"mol water"/"L water" = "55.348 mols"/"1 L" ~~ color(blue)("55.348 M")

EDIT:
The reason I give these 'absurd' concentrations is that what your teachers don't tell you (and this sounds like spam but it isn't!) is that since the concentration of water is so high as pure water, it isn't often discussed...

...until you need the "molar density" $\overline{\rho} = \frac{\rho}{n}$ of water in mol/L, which is just the molarity $\text{M}$ without the context of a solution, or the molar volume $\overline{V} = \frac{V}{n}$ in L/mol, which is the reciprocal of the "molar density".

The molar volume is often used in thermodynamics and liquid-liquid solution contexts, such as when calculating the freezing point depression using the full van't Hoff equation for ${K}_{f}$ ((""^@ "C")/"m").