At this point we will not limit ourselves to aqueous solutions, formed by water as a solvent and other minor components called solutes. Thus, when we write KCl(aq), we do not refer to an aqueous solution that contains KCl as a solute.

To work with these solutions, it is convenient to introduce new ways of establishing the ratio between solute and solvent, in addition to the percentage composition.

**Molarity** : It is defined as the ratio between the moles of solute and the volume of the solution. \begin{equation} Molarity\ (M)=\frac{Moles\ solute}{volume\ solution\acute{o}n\ (liters)} \end{equation} Molarity is represented by the symbol M and has units mol/L.

A 0.2 M (0.2 molar) solution of glucose contains 0.2 mol of glucose in 1 liter of solution.

**Example 1.** 15.0 mL of acetic acid, $CH_3COOH$ (d=1.048 g/mL), is dissolved in enough water to make 500.0 mL of solution. What is the molarity of the acetic acid in the solution?

Steps to follow: (1) Obtain the moles of acetic acid that are in 15 mL of solution; (2) divide the moles of acetic acid by 0.5 L. \begin{equation} 15\ mL\ AcOHx\frac{1.048\ g\ AcOH}{1\ mL\ AcOH}x\frac{1\ mol\ AcOH}{ 60.05\ g\ AcOH}=0.262\ mol\ AcOH \end{equation} \begin{equation} M=\frac{0.262}{0.500}=0.524\ mol/L \end{equation}