3.3 Molarity

Molarity and Solution Concentration

Molarity tells you how concentrated a solution is by measuring the moles of solute dissolved per liter of solution. It's one of the most common ways to describe concentration in chemistry, and you'll rely on it constantly for stoichiometric calculations and comparing solutions.

Concept of Molarity

Molarity (M) is defined as:

$M = \frac{\text{moles of solute}}{\text{liters of solution}}$

A solution with more moles of solute packed into the same volume has a higher molarity. For example, a 1.0 M sugar solution is ten times more concentrated than a 0.1 M sugar solution.

• Higher molarity = more concentrated; lower molarity = more dilute
• Molarity can change slightly with temperature because liquids expand when heated, increasing the volume of the solution. That's why molarity is typically reported at 25°C (room temperature).

Molarity Calculations and Unit Conversions

To calculate molarity, you need two things: moles of solute and volume of solution in liters. Here's the process:

1. Convert mass of solute to moles using the substance's molar mass (g/mol): $\text{Moles of solute} = \frac{\text{mass of solute}}{\text{molar mass of solute}}$
2. Convert solution volume to liters if it's given in other units (e.g., divide mL by 1000).
3. Divide moles by liters to get molarity.

Example: Find the molarity of a solution made by dissolving 25.0 g of NaCl in enough water to make 500 mL of solution.

1. Molar mass of NaCl = 58.44 g/mol
2. $\text{Moles of NaCl} = \frac{25.0\text{ g}}{58.44\text{ g/mol}} = 0.428\text{ mol}$
3. Volume = 500 mL = 0.500 L
4. $\text{Molarity} = \frac{0.428\text{ mol}}{0.500\text{ L}} = 0.856\text{ M}$

Solution Dilution and Concentration Problems

Dilution means adding more solvent to a solution to lower its concentration. The key idea is that the moles of solute don't change during dilution; you're just spreading them out in a larger volume. This gives us the dilution equation:

$M_1V_1 = M_2V_2$

where $M_1$ and $V_1$ are the initial molarity and volume, and $M_2$ and $V_2$ are the final molarity and volume.

To solve a dilution problem:

1. Identify which three values you know (two on one side, one on the other).
2. Make sure volumes are in the same units (both in mL or both in L).
3. Solve for the unknown using algebra.

Example: You take 100 mL of a 2.00 M KCl solution and dilute it to a total volume of 1.00 L. What's the final concentration?

1. $M_1 = 2.00\text{ M}$, $V_1 = 0.100\text{ L}$, $V_2 = 1.00\text{ L}$
2. $(2.00\text{ M})(0.100\text{ L}) = M_2(1.00\text{ L})$
3. $M_2 = \frac{(2.00)(0.100)}{1.00} = 0.200\text{ M}$

The concentration dropped by a factor of 10 because the volume increased by a factor of 10. That proportional relationship is always worth checking as a quick sanity test on your answer.

Solution Preparation and Concentration

Concentration is a general term for how much solute is dissolved in a given amount of solution. Molarity is one specific way to express concentration.

A solution is a homogeneous mixture, meaning the solute is evenly distributed throughout the solvent. When preparing a solution of known molarity in the lab, you'll typically use a volumetric flask, which is designed to hold a precise volume of liquid.

Stock solutions are pre-made concentrated solutions that you dilute down to whatever concentration you need for an experiment. This is where the $M_1V_1 = M_2V_2$ equation becomes especially useful: you calculate exactly how much stock solution to measure out, then add solvent to reach your target volume and concentration.