# Molarity of pure water [closed]

The molarity of pure water is 55.5 M at which temprature is obtained this value? Further, you will come across that 1000 mL = 1000 g (for water only, but at which temprature?)

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Water is very unique liquid, because it has a higher density in the liquid state than the solid state. The maximum density of water is found at $$\pu{4 ^\circ C}$$, which is reported as $$\pu{999.9720 kg\:m^{-3}}$$ (Temperature Effects on Density). Therefore, molarity of water at $$\pu{4 ^\circ C}$$ can be calculated as $$\frac{\pu{999.9720 g\:L^{-1}}}{\pu{18.015 g\:mol^{-1}}} = \pu{55.51 mol\:L^{-1}}$$.

Following chart shows accepted densities of pure water at range of temperatures:

$$\text{Table of Densities of Pure Water at Various temperatures under }\pu{1.0 atm}$$ $$\begin{array}{ccc} \hline \text{Temperature }(\pu{^\circ C }) &\text{Density }(\pu{g\:cm^{-3} }) & \text{Temperature }(\pu{^\circ C }) &\text{Density }(\pu{g\:cm^{-3} }) \\\hline -30 & 0.98385 & +30 & 0.99565 \\ -20 & 0.99355 & +35 & 0.99406 \\ -10 & 0.99812 & +40 & 0.99225 \\ -5 & 0.99930 & +45 & 0.99025 \\ 0 & 0.999840 & +50 & 0.98807 \\ +4 & 0.99997 & +60 & 0.98313 \\ +10 & 0.99970 & +70 & 0.97763 \\ +15 & 0.99910 & +80 & 0.97180 \\ +20 & 0.99821 & +90 & 0.96506 \\ +25 & 0.99705 & +100 & 0.95838 \\\hline \end{array}$$

So you may calculate the molarity of pure water at each temperature using the chart. An example for step by step calculations to find the molarity of pure water at particular temperature ($$\pu{20 ^\circ C}$$):

$$\text{Density of pure water at }\pu{20 ^\circ C } = \pu{0.99821 g\:cm^{-3}}$$ Since molarity $$= \pu{mol\:L^{-1} }$$, let's consider $$\pu{1.0 L}$$ of water. $$\text{Mass of } \pu{1.0 L} \text{ of pure water at }\pu{20 ^\circ C} = \pu{1.0 L}\times \frac{\pu{1000 cm^{-3}}}{\pu{1.0 L}} \times \pu{0.99821 g\:cm^{-3}}= \pu{998.21 g}$$ $$\text{Molar mass of pure water at any temperature} = \pu{18.015 g\:mol^{-1}}$$ $$\therefore \text{ # of moles in } \pu{1.0 L} \text{ of pure water at }\pu{20 ^\circ C} = \frac{\pu{998.21 g}}{\pu{18.015 g\:mol^{-1}}} = \pu{55.41 mol}$$ $$\therefore \text{Molarity of pure water at } \pu{20 ^\circ C} = \frac{\pu{55.41 mol}}{\pu{1.0 L}} = \pu{55.41 mol\:L^{-1}}$$

• Problem: undefined control sequence\pu...this statement is repeated 4 times in your answer, what to do? – yaseen wazir Jun 2 at 5:24
• Please don’t post text and equations as a picture. The content cannot be found by search engines and screen readers. It makes it difficult for other users to edit your answer. Especially since this answer is directed at beginners, you might want to consider using proper equations and typography. Please do not use names of quantities and names of units in equations. And please use correct units symbols. – Loong Jun 2 at 7:34
• @yaseenwazir Please refer to Undefined control sequence \pu in official SE Android app. Note that mobile Stack Exchange app is using outdated MathJax and hasn't been updated for a long time. I strongly recommend to use a mobile web browser instead. – andselisk Jun 2 at 16:43
• @Mithoron Mathew is doing it right, or at least according to the way it should be done on Chemistry.SE, see How to write physical units. – andselisk Jun 2 at 16:46
• @Loong: Thanks for the valuable point. I apologize the community for been lazy. I found this post in a website in PEG format, but so sleepy to write it up and decided to attach as a picture. Won't happen again. – Mathew Mahindaratne Jun 2 at 23:27