Problem 36
Question
What volume of a 0.33- \(M\) C \(_{12} \mathrm{H}_{22} \mathrm{O}_{11}\) solution can be diluted to prepare 25 \(\mathrm{mL}\) of a solution with a concentration of \(0.025 \mathrm{M}\) ?
Step-by-Step Solution
Verified Answer
Approximately 1.89 mL of the 0.33 M C12H22O11 solution is needed to prepare 25 mL of a 0.025 M solution.
1Step 1: Determine the information given and what is needed
We have the initial concentration (M1) of the C12H22O11 solution, which is 0.33 M, and the final concentration (M2) we want after dilution, which is 0.025 M. We also know the final volume (V2) needed, which is 25 mL. Our goal is to find the initial volume (V1) of the 0.33 M solution needed to make this dilution.
2Step 2: Use the dilution equation
The dilution equation is M1 * V1 = M2 * V2. We will use this equation to solve for V1.
3Step 3: Rearrange the dilution equation to solve for V1
To find V1, rearrange the dilution equation to V1 = (M2 * V2) / M1.
4Step 4: Plug in the values and calculate V1
Now that we have the dilution equation arranged to solve for V1, we can plug in the values we know. V1 = (0.025 M * 25 mL) / 0.33 M.
5Step 5: Find the initial volume (V1)
Calculate the value of V1 using the values from step 4. V1 = (0.025 * 25) / 0.33 = 0.625 / 0.33 = 1.89393939 mL (approximated to 1.89 mL).
Key Concepts
MolaritySolution ConcentrationChemistry CalculationsVolume Dilution
Molarity
Molarity (\( M \)) is a measure of the concentration of a solute in a solution, or how many moles of a substance are in a liter of solution. It's calculated using the formula:
\[ M = \frac{\text{moles of solute}}{\text{liters of solution}} \]
In the context of the given exercise, we are looking at a sugar (\( C_{12}H_{22}O_{11} \) solution. Molarity would allow us to determine how much sugar is present in a specific volume of the liquid. Molarity is fundamental in chemistry calculations because it provides a standardized way to describe solution concentration which ensures consistency when conducting experiments or comparing findings.
\[ M = \frac{\text{moles of solute}}{\text{liters of solution}} \]
In the context of the given exercise, we are looking at a sugar (\( C_{12}H_{22}O_{11} \) solution. Molarity would allow us to determine how much sugar is present in a specific volume of the liquid. Molarity is fundamental in chemistry calculations because it provides a standardized way to describe solution concentration which ensures consistency when conducting experiments or comparing findings.
Solution Concentration
Solution concentration refers to the amount of solute that is dissolved in a given quantity of solvent. The concentration can be expressed in various ways, such as molarity, mass percent, or parts per million (ppm). In our exercise, we use molarity to indicate concentration, which makes comparisons and calculations straightforward. Understanding solution concentration is critical for predicting the behavior of solutions, such as how they react in a chemical reaction or how they conduct electricity.
Numerical Expressions of Concentration
When a solution is more concentrated, it means there is a higher amount of solute compared to solvent. Conversely, a dilute solution has a relatively low amount of solute. Solution concentration can greatly affect the rate of chemical reactions and the physical properties of the solution, such as boiling point and freezing point.Chemistry Calculations
Chemistry calculations often involve using mathematical expressions to understand and predict the behavior of chemical substances. Calculations like the one in our exercise are essential for preparing solutions of a desired concentration. They can also extend to determining reaction yields, calculating pKa values for acids and bases, and much more.
Importance of Accuracy
Accurate chemistry calculations are crucial in scientific research, consistently producing desired reactions, and ensuring safety in experiments. When solving for volume using the molarity equation, it's important to perform conversion when necessary (e.g., milliliters to liters) to maintain unit consistency and precision.Volume Dilution
Volume dilution is the process of reducing the concentration of a solution by adding more solvent to it. The dilution equation, \[ M1 \times V1 = M2 \times V2 \]
relates the molarity (M) and volume (V) of the initial solution to the molarity and volume of the final, diluted solution. In our exercise, we found the initial volume required to produce a diluted solution with a known final molarity and volume.
relates the molarity (M) and volume (V) of the initial solution to the molarity and volume of the final, diluted solution. In our exercise, we found the initial volume required to produce a diluted solution with a known final molarity and volume.
Applications of Volume Dilution
Understanding volume dilution is essential in many fields like pharmaceuticals, where medications need to be diluted to safe concentrations, or in biochemistry, where the activity of enzymes can depend on the concentration of substrates. Volume dilution enables chemists to prepare solutions of precise concentrations for their experimental needs or commercial products.Other exercises in this chapter
Problem 34
If 0.1718 L of a \(0.3556-M C_{3} \mathrm{H}_{7}\) OH solution is diluted to a concentration of \(0.1222 M,\) what is the volume of the resulting solution?
View solution Problem 35
If 4.12 L of a 0.850 \(M-\mathrm{H}_{3} \mathrm{PO}_{4}\) solution is be diluted to a volume of 10.00 \(\mathrm{L},\) what is the concentration of the resulting
View solution Problem 37
What is the concentration of the NaCl solution that results when 0.150 L of a 0.556-M solution is allowed to evaporate until the volume is reduced to \(0.105 \m
View solution Problem 38
What is the molarity of the diluted solution when each of the following solutions is diluted to the given final volume? (a) \(1.00 \mathrm{L}\) of a \(0.250-M\)
View solution