91Ó°ÊÓ

Dilution is a crucial concept in chemistry that allows us to reduce the concentration of a solution by adding more solvent. The key formula used in dilution calculations is:\[ C_1 \times V_1 = C_2 \times V_2 \]Here:

• \( C_1 \) is the initial concentration
• \( V_1 \) is the initial volume
• \( C_2 \) is the final concentration
• \( V_2 \) is the final volume

Using this equation, you can determine how the concentration changes as you adjust the volume. In our exercise, solutions A, B, and C were prepared by sequentially diluting a stock solution. For each step:

• First, we calculate the number of moles in the smaller volume (\( V_1 \)).
• Then, divide by the new total volume (\( V_2 \)) to find the new concentration (\( C_2 \)).

Understanding this process helps predict the concentrations at each stage, making it easier to plan experiments or processes that require specific solution strengths.

The mole concept is fundamental in chemistry, allowing us to quantify amounts in chemical reactions. A mole represents \( 6.022 \times 10^{23} \) entities (Avogadro's number) of any substance.In the context of our problem, we first determined how many moles of manganese were present in the stock solution. This involves the formula:\[ \text{number of moles} = \frac{\text{mass}}{\text{molar mass}} \]For manganese, we used its molar mass found in the periodic table, which is \( 54.94 \text{ g/mol} \). By dividing the mass of manganese used \( (1.584 \text{ g}) \) by its molar mass, we found the number of moles.This foundational step is essential because the number of moles will determine the initial concentration in the stock solution and guide subsequent dilution calculations. It tells us how much substance is actually present, regardless of any chemical formula, making it a versatile tool for chemists.

Molarity is a measure of the concentration of a solution, expressed in moles of solute per liter of solution. In our problem, we calculated the molarity of the initial stock solution and then proceeded with solutions A, B, and C.Start with the formula:\[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \]For the stock solution, we used the moles calculated earlier and divided by the total volume (1.000 L). This gave us the molarity of the starting solution. Each subsequent dilution followed the same principle but adjusted for the new volumes.Molarity is significant because it allows chemists to easily relate volumes and concentrations in a practical way, crucial for mixing solutions accurately. This concept underpins many aspects of laboratory work, providing a clear measure of how solute is distributed in a solution.