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Chapter 15: Problem 123

How many milliliters of \(18.0 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) are required to prepare \(35.0 \mathrm{~mL}\) of \(0.250 \mathrm{M}\) solution?

Short Answer

Expert verified
Approximately 0.486 mL of 18.0 M \(H_{2}SO_{4}\) are required to prepare 35.0 mL of 0.250 M solution.

Step by step solution

01

Identify the given information

We are given the following information: Initial concentration, \(C_{1} = 18.0~M\) Final concentration, \(C_{2} = 0.250~M\) Final volume, \(V_{2} = 35.0~mL\) We need to find the initial volume, \(V_{1}\).
02

Use the dilution equation

We will use the dilution equation to find the initial volume, \(V_{1}\): \(C_{1}V_{1}= C_{2}V_{2}\) Plug in the given information: \(18.0~M * V_{1} = 0.250~M * 35.0~mL\)
03

Solve for the initial volume \(V_{1}\)

Now, we will solve for \(V_{1}\): \(V_{1} = \frac{0.250~M * 35.0~mL}{18.0~M}\) Calculate the result: \(V_{1} \approx 0.486~mL\)
04

State the answer

Approximately 0.486 mL of 18.0 M \(H_{2}SO_{4}\) are required to prepare 35.0 mL of 0.250 M solution.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Molarity
Molarity is a measure of the concentration of a solute in a solution, expressed as the number of moles of solute per liter of solution. It's a critical concept in chemistry for understanding the strength of solutions:
  • Molarity (M) indicates the number of moles of solute compared to the volume (in liters) of the solution.
  • The formula for molarity is given by: \[ M = \frac{n}{V} \]where \( n \) is the number of moles of solute, and \( V \) is the volume of the solution in liters.
  • For example, a solution labeled as 18.0 M means it contains 18.0 moles of solute per liter.
Knowing molarity is essential for tasks like mixing solutions or reacting chemicals where quantities and reactions depend on how concentrated each component is. It's particularly useful in lab settings and industrial applications, where precise concentrations are crucial for successful reactions.
Solution Preparation
When preparing a solution, it is important to precisely calculate the amounts needed to achieve the desired concentration. Here's the process:
  • Determine the desired molarity and volume: Decide how concentrated your solution needs to be and the total volume required. This information is crucial to know before beginning.
  • Calculate the mass or volume of solute required: Use the formula for molarity to determine how much solute is needed. For a strong acid like sulfuric acid (\( H_2SO_4 \)), you'll measure a specific volume if concentrated liquid is used directly.
  • Steps in the preparation include measuring the solvent (often water) and then adding the solute while mixing to ensure complete dissolution. After adding, continue mixing until you achieve uniform concentration throughout the solution.
Always remember to adjust temperatures and handle chemicals safely, as precision is key to achieving the perfect solution for experiments or applications.
Dilution Equation
The dilution equation is a valuable tool used in preparing solutions of lower concentrations from a more concentrated stock solution. This equation is:\[ C_1V_1 = C_2V_2 \]
  • Understanding the variables:
    • \( C_1 \): initial concentration
    • \( V_1 \): initial volume
    • \( C_2 \): final concentration desired
    • \( V_2 \): final volume needed
    This equation shows that the product of the initial concentration and volume (\( C_1V_1 \)) must equal the product of the final desired concentration and volume (\( C_2V_2 \)).
  • Practical application: The dilution equation helps determine how much of a concentrated solution is needed to reach a desired concentration. For example, if you have a very concentrated acid and need a specific lower concentration, you can use the dilution equation to calculate exactly how much of the concentrated acid to dilute.
By applying this equation, you can ensure accurate, consistent results, whether for classroom experiments, laboratory research, or industrial applications.

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Most popular questions from this chapter

Calculate the number of moles of each ion present in each of the following solutions. a. \(10.2 \mathrm{~mL}\) of \(0.451 \mathrm{M} \mathrm{AlCl}_{3}\) solution b. \(5.51 \mathrm{~L}\) of \(0.103 \mathrm{M} \mathrm{Na}_{3} \mathrm{PO}_{4}\) solution c. \(1.75 \mathrm{~mL}\) of \(1.25 \mathrm{M} \mathrm{CuCl}_{2}\) solution d. \(25.2 \mathrm{~mL}\) of \(0.00157 \mathrm{M} \mathrm{Ca}(\mathrm{OH})_{2}\) solution

What \(mass\) of the indicated solute does each of the following solutions contain? a. \(2.50 \mathrm{~L}\) of \(13.1 \mathrm{M} \mathrm{HCl}\) solution b. \(15.6 \mathrm{~mL}\) of \(0.155 \mathrm{M} \mathrm{NaOH}\) solution c. \(135 \mathrm{~mL}\) of \(2.01 \mathrm{M} \mathrm{HNO}_{3}\) solution d. 4.21 L of \(0.515 M \mathrm{CaCl}_{2}\) solution

A solution is prepared by dissolving \(0.6706 \mathrm{~g}\) of oxalic acid \(\left(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\right)\) in enough water to make \(100.0 \mathrm{~mL}\) of solution. \(\mathrm{A} 10.00-\mathrm{mL}\) aliquot (portion) of this solution is then diluted to a final volume of \(250.0 \mathrm{~mL}\). What is the final molarity of the oxalic acid solution?

Concentrated hydrochloric acid is made by pumping hydrogen chloride gas into distilled water. If concentrated \(\mathrm{HCl}\) contains \(439 \mathrm{~g}\) of \(\mathrm{HCl}\) per liter, what is the molarity?

For each of the following solutions, the mass of solute taken is indicated, as well as the total volume of solution prepared. Calculate the normality of each solution. a. \(15.0 \mathrm{~g}\) of \(\mathrm{HCl} ; 500 . \mathrm{mL}\) b. \(49.0 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{SO}_{4} ; 250 . \mathrm{mL}\) c. \(10.0 \mathrm{~g}\) of \(\mathrm{H}_{3} \mathrm{PO}_{4} ; 100 . \mathrm{mL}\)
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