Understanding Molarity

Molarity is a key concept in chemistry that describes the concentration of a solution. It is defined as the number of moles of solute per liter of solution. Understanding how to calculate molarity from mass and volume is crucial for preparing solutions accurately in laboratory settings. Below, we present three diverse examples that illustrate how to perform these calculations in practical contexts.

Example 1: Preparing a Sodium Chloride Solution

In a laboratory setting, you may need to prepare a sodium chloride (NaCl) solution for an experiment. For this example, let’s say you want to create a 0.5 M solution of NaCl in 500 mL of water.

To find out how much NaCl you need:

• First, calculate the number of moles required for the desired molarity:

  • Molarity (M) = Moles of solute / Volume of solution in liters
  • Rearranging gives us: Moles of solute = Molarity × Volume
  • Moles of NaCl = 0.5 M × 0.5 L = 0.25 moles

• Next, we need to convert moles to grams. The molar mass of NaCl is approximately 58.44 g/mol:

  • Mass of NaCl = Moles × Molar Mass
  • Mass of NaCl = 0.25 moles × 58.44 g/mol = 14.61 g

• Therefore, you will need to weigh out approximately 14.61 grams of NaCl and dissolve it in enough water to make a total volume of 500 mL.

Notes

• If you want to prepare a different volume or concentration, simply adjust the values accordingly.

Example 2: Diluting Acetic Acid for a Laboratory Experiment

Suppose you have a concentrated acetic acid solution (CH₃COOH) with a molarity of 6 M, and you need to prepare 250 mL of a 1 M solution for a laboratory experiment.

To find out how much of the concentrated solution you need:

• First, use the dilution formula (C₁V₁ = C₂V₂), where C₁ is the concentration of the stock solution, V₁ is the volume of the stock solution needed, C₂ is the final concentration, and V₂ is the final volume:
  • 6 M × V₁ = 1 M × 0.25 L

• Now, solve for V₁:

  • V₁ = (1 M × 0.25 L) / 6 M = 0.04167 L or 41.67 mL

• Therefore, you will need to measure out 41.67 mL of the 6 M acetic acid and dilute it with distilled water to a total volume of 250 mL.

Notes

• Always add acid to water, not water to acid, to avoid exothermic reactions that can cause splattering.

Example 3: Calculating Molarity from a Given Mass of Potassium Nitrate

In agricultural applications, potassium nitrate (KNO₃) is often used as a fertilizer. Let’s say you want to prepare a 2 M solution of KNO₃ using 100 grams of the compound. First, we need to calculate how many moles of KNO₃ are in 100 grams:

• The molar mass of KNO₃ is approximately 101.1 g/mol.

• Moles of KNO₃ = Mass / Molar Mass

  • Moles of KNO₃ = 100 g / 101.1 g/mol ≈ 0.990 moles

• Now, to find the volume of solution needed to achieve a 2 M concentration:

  • Molarity = Moles of solute / Volume of solution in liters
  • Volume = Moles of solute / Molarity
  • Volume = 0.990 moles / 2 M = 0.495 L or 495 mL

• Therefore, you would need to dissolve 100 grams of KNO₃ in enough water to make a total volume of 495 mL to achieve a 2 M solution.

Notes

• Ensure that the KNO₃ is fully dissolved before measuring the final volume to avoid discrepancies in concentration.