Understanding Molarity

Molarity (M) is a key concept in chemistry that quantifies the concentration of a solution. It is defined as the number of moles of solute per liter of solution. Molarity is essential for various applications, including chemical reactions, solution preparation, and laboratory experiments. Here, we present three diverse examples to illustrate how to determine the concentration of a solution using molarity.

Example 1: Preparing a Sodium Chloride Solution

In a laboratory, a chemist needs to prepare a 0.5 M sodium chloride (NaCl) solution for an experiment. The goal is to understand how to calculate the amount of NaCl needed to achieve this molarity.

1. Calculate Molar Mass: The molar mass of NaCl is approximately 58.44 g/mol (Na: 22.99 g/mol + Cl: 35.45 g/mol).
2. Determine Volume of Solution: The chemist decides to prepare 1 liter of the solution.

3. Calculate Moles Needed:

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

   0.5 M = Moles / 1 L

   Moles = 0.5 moles

4. Convert Moles to Grams:

   Grams = Moles × Molar Mass

   Grams = 0.5 moles × 58.44 g/mol = 29.22 g

To prepare the 0.5 M NaCl solution, the chemist dissolves 29.22 grams of sodium chloride in enough water to make a final volume of 1 liter.

Notes:

• Ensure the sodium chloride is completely dissolved before measuring the final volume.
• Adjustments might be necessary for temperature and pressure effects on solubility.

Example 2: Analyzing a Sugar Solution

A nutritionist is studying the concentration of glucose in a beverage. The beverage contains 15 grams of glucose (C6H12O6) dissolved in 250 mL of water. The nutritionist needs to determine the molarity of the glucose solution.

1. Calculate Molar Mass: The molar mass of glucose is approximately 180.18 g/mol.

2. Convert grams to moles:

   Moles = Grams / Molar Mass

   Moles = 15 g / 180.18 g/mol ≈ 0.0833 moles

3. Convert volume to liters:

   250 mL = 0.250 L

4. Calculate Molarity:

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

   Molarity = 0.0833 moles / 0.250 L = 0.333 M

The glucose solution has a molarity of approximately 0.333 M. This information is crucial for understanding the nutritional content of the beverage.

Notes:

• Molarity can vary with temperature; always check the temperature of the solution.
• This calculation can help in dietary planning and understanding glucose levels in foods.

Example 3: Diluting a Concentrated Acid

A laboratory technician needs to dilute 6 M hydrochloric acid (HCl) to a 1 M solution for safety and ease of use. The technician wants to know how much of the concentrated solution is required to make 2 liters of the diluted solution.

1. Use the dilution formula:

   C1V1 = C2V2

   Where:

   • C1 = concentration of the concentrated solution (6 M)
   • V1 = volume of concentrated solution needed (unknown)
   • C2 = concentration of the diluted solution (1 M)
   • V2 = volume of the diluted solution (2 L)

2. Plug in the values:

   6 M × V1 = 1 M × 2 L

3. Solve for V1:

   V1 = (1 M × 2 L) / 6 M = 0.333 L = 333 mL

To create a 1 M hydrochloric acid solution, the technician must measure 333 mL of the 6 M solution and dilute it with distilled water to a final volume of 2 liters.

Notes:

• Always add acid to water, not water to acid, to prevent exothermic reactions that can cause splattering.
• Proper safety gear, including gloves and goggles, should be worn during this process.