Understanding Molarity Calculations

Molarity (M) is a fundamental concept in chemistry that quantifies the concentration of a solute in a solution. It is defined as the number of moles of solute per liter of solution. Molarity is essential for various applications, including reaction stoichiometry, solution preparation, and analytical chemistry. In this article, we will explore three diverse practical examples of molarity calculations to enhance your understanding.

Example 1: Preparing a Sodium Chloride Solution

In laboratory settings, chemists often need to prepare solutions of specific concentrations. For this example, we will prepare a sodium chloride (NaCl) solution with a molarity of 0.5 M.

To prepare 1 liter of a 0.5 M NaCl solution, we first need to calculate the number of moles of NaCl required. The molar mass of NaCl is approximately 58.44 g/mol.

1. Calculate the moles required:

   Molarity (M) = Moles of solute / Volume of solution (L)

   Rearranging gives us:

   Moles of solute = Molarity × Volume = 0.5 M × 1 L = 0.5 moles

2. Convert moles to grams:

   Mass (g) = Moles × Molar Mass = 0.5 moles × 58.44 g/mol = 29.22 g

3. Weigh out 29.22 grams of NaCl and dissolve it in enough water to make a total volume of 1 liter.

Notes:

• If you need to prepare a different volume, simply adjust the volume in the calculations accordingly.
• Ensure to mix the solution thoroughly to achieve uniformity.

Example 2: Diluting a Stock Solution

This example illustrates how to dilute a concentrated stock solution to achieve a desired molarity. Assume you have a stock solution of hydrochloric acid (HCl) with a molarity of 6 M and you want to dilute it to 2 M.

1. Use the dilution equation:

   C1V1 = C2V2

   Where:

   • C1 = initial concentration (6 M)
   • V1 = volume of stock solution needed
   • C2 = final concentration (2 M)
   • V2 = final volume of diluted solution (let’s say 1 L)

2. Rearranging the equation to find V1:

   V1 = (C2 × V2) / C1 = (2 M × 1 L) / 6 M = 0.333 L (or 333.33 mL)

3. Measure 333.33 mL of the 6 M HCl stock solution and add enough water to bring the total volume to 1 liter.

Notes:

• Always add acid to water, not the other way around, to ensure safety.
• The dilution factor can be useful for calculating concentrations in various scenarios.

Example 3: Calculating Molarity from Mass and Volume

In this example, we will calculate the molarity of a solution when given the mass of solute and the volume of the solution. Suppose we have 50 grams of potassium nitrate (KNO3) dissolved in enough water to make a total volume of 500 mL.

1. First, calculate the molar mass of KNO3. The molar mass is approximately 101.10 g/mol.

2. Convert grams to moles:

   Moles = Mass (g) / Molar Mass (g/mol) = 50 g / 101.10 g/mol ≈ 0.495 moles

3. Convert the volume from milliliters to liters:

   Volume (L) = 500 mL / 1000 = 0.5 L

4. Now calculate the molarity:

   Molarity (M) = Moles of solute / Volume of solution (L) = 0.495 moles / 0.5 L = 0.99 M

Notes:

• This method is highly useful for determining concentrations in various chemical processes.
• Always ensure accurate measurements for precise calculations.