Problem 48
Question
If \(4.00 \mathrm{mL}\) of \(0.0250 \mathrm{M} \mathrm{CuSO}_{4}\) is diluted to \(10.0 \mathrm{mL}\) with pure water, what is the molar concentration of copper(II) sulfate in the diluted solution?
Step-by-Step Solution
Verified Answer
The molar concentration of copper(II) sulfate in the diluted solution is 0.0100 M.
1Step 1: Understand Dilution Formula
The dilution formula is given by: \[ C_1V_1 = C_2V_2 \] where \( C_1 \) is the initial concentration, \( V_1 \) is the initial volume, \( C_2 \) is the final concentration, and \( V_2 \) is the final volume.
2Step 2: Identify Given Values
We know that \( C_1 = 0.0250 \mathrm{M} \), \( V_1 = 4.00 \mathrm{mL} \), and \( V_2 = 10.0 \mathrm{mL} \). We need to find \( C_2 \).
3Step 3: Rearrange Dilution Formula
Rearrange the formula to solve for \( C_2 \): \[ C_2 = \frac{C_1V_1}{V_2} \]
4Step 4: Substitute Values
Substitute the known values into the equation: \[ C_2 = \frac{0.0250 \times 4.00}{10.0} \]
5Step 5: Calculate Final Concentration
Perform the calculation: \[ C_2 = \frac{0.100}{10.0} = 0.0100 \mathrm{M} \].
Key Concepts
Molar ConcentrationCopper(II) SulfateSolution Dilution
Molar Concentration
Molar concentration, also known as molarity, is a measure of the concentration of a solute in a solution. It tells us how many moles of a substance are present in one liter of solution. The unit used for molarity is molar, represented by "M".
To calculate molarity, you can use the formula: \[M = \frac{n}{V}\]where:
To calculate molarity, you can use the formula: \[M = \frac{n}{V}\]where:
- \(M\) is the molarity in moles per liter (M)
- \(n\) is the number of moles of solute
- \(V\) is the volume of the solution in liters
Copper(II) Sulfate
Copper(II) sulfate, commonly known by its chemical formula \(\text{CuSO}_4\), is a popular chemical compound in labs and industrial applications. It is known for its distinctive blue color when hydrated, as it commonly exists as a pentahydrate \(\text{CuSO}_4\cdot5\text{H}_2\text{O}\). This is the form that appears most often in chemistry laboratories.
Here are several important aspects of Copper(II) sulfate to understand:
Here are several important aspects of Copper(II) sulfate to understand:
- It is a highly soluble salt in water, making it ideal for various solutions.
- Frequently used in agriculture and industries for activities such as pest control and electroplating.
- In educational settings, it's often used to demonstrate chemical reactions and properties, such as conducting specific precipitation reactions.
Solution Dilution
Solution dilution is a process whereby the concentration of a solute in a solution is decreased by adding more solvent. It helps chemists and students create solutions of desired concentrations for experiments and various applications.
The fundamental principle behind dilution is that the number of moles of solute remains constant before and after dilution. Consequently, the dilution formula \(C_1V_1 = C_2V_2\) is used to calculate how much solvent is needed or to find the new concentration.
The formula is interpreted as:
The fundamental principle behind dilution is that the number of moles of solute remains constant before and after dilution. Consequently, the dilution formula \(C_1V_1 = C_2V_2\) is used to calculate how much solvent is needed or to find the new concentration.
The formula is interpreted as:
- \(C_1\) and \(V_1\) are the initial concentration and volume, respectively.
- \(C_2\) is the final concentration we want to find, and \(V_2\) is the final volume after dilution.
Other exercises in this chapter
Problem 46
What mass of oxalic acid, \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4},\) is required to prepare \(250 .\) mL of a solution that has a concentration of \(0.15
View solution Problem 47
If you dilute \(25.0 \mathrm{mL}\) of \(1.50 \mathrm{M}\) hydrochloric acid to \(500 . \mathrm{mL},\) what is the molar concentration of the dilute acid?
View solution Problem 49
Which of the following methods would you use to prepare \(1.00 \mathrm{L}\) of \(0.125 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4} ?\) (a) Dilute \(20.8 \mathrm{m
View solution Problem 50
Which of the following methods would you use to prepare \(300 .\) mL. of \(0.500 \mathrm{M} \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} ?\) (a) Add \(30.0 \ma
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