Dilution is a common process in chemistry where a concentrated solution is mixed with a solvent to decrease its concentration. Molarity, a key concept in solution chemistry, is defined as the number of moles of solute per liter of solution. This article provides three practical examples of dilution and molarity calculations that illustrate how these concepts are applied in real-world scenarios.
In a laboratory setting, a chemist needs to prepare a 0.5 M sodium chloride (NaCl) solution from a stock solution that has a concentration of 2 M. This situation is common for experiments requiring specific concentrations of reagents.
To find out how much of the stock solution is needed, the dilution formula can be used:
C1V1 = C2V2
Where:
C1 = concentration of the stock solution (2 M)
V1 = volume of the stock solution needed
C2 = concentration of the diluted solution (0.5 M)
V2 = final volume of the diluted solution (1 L)
Solving for V1:
2 M * V1 = 0.5 M * 1 L
V1 = (0.5 M * 1 L) / 2 M
V1 = 0.25 L or 250 mL
To prepare the solution, the chemist would measure 250 mL of the 2 M NaCl solution and dilute it with distilled water to a final volume of 1 L. This example illustrates how dilution allows for the creation of solutions with desired concentrations for various laboratory applications.
A food scientist is analyzing a sugar (sucrose) solution to determine its molarity. The scientist dissolves 342 grams of sucrose in enough water to make 2 liters of solution. This calculation is essential for quality control in food production.
First, calculate the number of moles of sucrose:
Molar mass of sucrose (C12H22O11) = 342 g/mol
Moles of sucrose = mass (g) / molar mass (g/mol)
Moles of sucrose = 342 g / 342 g/mol = 1 mol
Next, calculate the molarity:
Molarity (M) = moles of solute / liters of solution
Molarity = 1 mol / 2 L = 0.5 M
This example highlights the importance of molarity in food science, where the concentration of ingredients affects taste and texture. By knowing the molarity of the sucrose solution, the scientist can make informed decisions about product formulation.
In an educational chemistry lab, a teacher wants to demonstrate pH testing using a diluted hydrochloric acid (HCl) solution. The teacher starts with a concentrated HCl solution at a concentration of 12 M and needs to prepare 500 mL of a 1 M solution for the demonstration.
Using the dilution formula:
C1V1 = C2V2
Where:
C1 = 12 M (initial concentration)
V1 = volume of concentrated solution needed
C2 = 1 M (final concentration)
V2 = 0.5 L (final volume)
Solving for V1:
12 M * V1 = 1 M * 0.5 L
V1 = (1 M * 0.5 L) / 12 M
V1 = 0.04167 L or 41.67 mL
The teacher would measure 41.67 mL of the concentrated HCl and dilute it with distilled water to make a total volume of 500 mL. This example is particularly useful for illustrating the concept of pH and the effects of acid concentration in a classroom setting. Understanding dilution and molarity calculations is crucial for safe laboratory practices and effective teaching.