3.3 Molarity

Molarity measures how concentrated a solution is by calculating moles of solute per liter of solution. It's crucial for comparing solutions and doing chemistry calculations. Knowing molarity helps you understand how much stuff is dissolved in a liquid.

You can calculate molarity by finding the moles of solute and volume of solution in liters. This involves converting mass to moles and volume to liters. Dilution problems use the equation M1V1 = M2V2 to figure out new concentrations when adding more solvent.

Molarity and Solution Concentration

Concept of molarity

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• Measures concentration of solutions by calculating moles of solute per liter of solution
  • Expressed as $\frac{\text{moles of solute}}{\text{liters of solution}}$
  • Commonly used in chemistry to quantify solution concentration (NaCl in water)
• Enables comparison of different solution concentrations and facilitates stoichiometric calculations
  • Higher molarity signifies more concentrated solution, lower molarity indicates more dilute solution (0.1 M vs 1.0 M sugar solution)
• Varies with temperature due to changes in solution volume
  • Typically reported at standard temperature, often 25°C (room temperature)
  • Molarity decreases with increasing temperature as volume expands (heating 1.0 M solution)

Molarity calculations and unit conversions

• Calculate molarity by determining moles of solute and volume of solution in liters
  1. Convert mass of solute to moles using molar mass
     • Molar mass represents mass of one mole of substance, usually in grams per mole (g/mol)
     • $\text{Moles of solute} = \frac{\text{mass of solute}}{\text{molar mass of solute}}$
  2. Convert solution volume to liters if needed
     • Apply conversion factors to convert from other units to liters (mL to L)
• Example: Find molarity of solution with 25.0 g NaCl in 500 mL 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 of solution = 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 involves adding more solvent to decrease solution concentration
  • Moles of solute stays constant, but solution volume increases
  • $M_1V_1 = M_2V_2$, where $M_1$, $V_1$ are initial molarity and volume, $M_2$, $V_2$ are final molarity and volume
• Solve dilution problems using $M_1V_1 = M_2V_2$ equation
  1. Identify known values (initial molarity, volume, final molarity or volume)
  2. Solve for unknown value using algebra
• Example: 2.00 M KCl solution diluted to 1.00 L using 100 mL of original solution. Find final concentration.
  • $M_1 = 2.00\text{ M}$, $V_1 = 100\text{ mL} = 0.100\text{ L}$, $V_2 = 1.00\text{ L}$
  • $M_1V_1 = M_2V_2$
  • $(2.00\text{ M})(0.100\text{ L}) = M_2(1.00\text{ L})$
  • $M_2 = \frac{(2.00\text{ M})(0.100\text{ L})}{1.00\text{ L}} = 0.200\text{ M}$

Solution Preparation and Concentration

• Concentration refers to the amount of solute dissolved in a given amount of solvent or solution
• A solution is a homogeneous mixture of two or more substances
• Volumetric flasks are used to accurately prepare solutions of known concentration
• Stock solutions are concentrated solutions used to prepare more dilute solutions for experiments