# What is the mol fraction of water in an aqueous solution of "3 M NaOH"?

May 4, 2017

Assuming $\text{298.15 K}$, the density of water is $\text{0.9970749 g/mL}$, or $\text{997.0749 g/L}$. Besides that, we need the mols of water in the solution before we can find its mol fraction.

We approximate that the density of the solution is that of water, and assume a $\text{1-L}$ solution:

$\text{1 L soln" xx "997.0749 g water"/"1 L soln" xx "1 mol water"/"18.015 g" ~~ "55.35 mols water}$

From this we can tell that this is dilute enough for the above approximation. Proceeding, we can then get the mol fraction. To illustrate a point, let us first find the mol fraction of solute:

chi_(NaOH) = "3 mols NaOH"/("3 mols NaOH" + "55.35 mols water")

$= 0.0514$

Therefore, the mol fraction of water is approximately $\textcolor{b l u e}{0.9486}$.