Molarity of pure water
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: 

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

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