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Chapter 2: Problem 122

\(100.0 \mathrm{~mL}\) of an \(18.0 \mathrm{M}\) solution of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) are transferred to a flask and diluted to \(500.0 \mathrm{~mL}\). What is the concentration of this new solution?

Short Answer

Expert verified
The new concentration is 3.6 M.

Step by step solution

01

Understand the Concept

This exercise involves diluting a concentrated solution to find its new concentration. The key principle here is that the amount of solute before dilution is equal to the amount of solute after dilution. This is represented by the formula \( C_1V_1 = C_2V_2 \), where \( C \) is concentration and \( V \) is volume.
02

Identify Known Values

From the problem, we know that the initial concentration \( C_1 = 18.0 \text{ M} \) and the initial volume \( V_1 = 100.0 \text{ mL} \). The final volume \( V_2 = 500.0 \text{ mL} \). We need to find the final concentration \( C_2 \).
03

Apply Dilution Formula

Use the formula \( C_1V_1 = C_2V_2 \). Substitute in the known values to get:\[ 18.0 \times 100.0 = C_2 \times 500.0 \]
04

Solve for Final Concentration

Rearrange the equation from Step 3 to solve for \( C_2 \):\[ C_2 = \frac{18.0 \times 100.0}{500.0} \]Simplify the calculation:\[ C_2 = \frac{1800.0}{500.0} = 3.6 \text{ M} \]

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Key Concepts

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

Concentration
Concentration refers to the amount of a substance, known as the solute, that is present in a given volume of a solution. It is typically expressed in moles per liter (Molarity, denoted as "M"). In this context, a higher concentration means more solute is packed into the solution.
  • Molarity (M) is calculated as the moles of solute divided by the liters of solution.
  • In our example, the initial concentration of \(\mathrm{{H}_{2} \mathrm{SO}_{4}}\) is 18.0 M, which means there are 18 moles of sulfuric acid in one liter of solution.
Concentration is crucial as it helps determine how much of the solute is active in reactions. It's the starting point for calculations in dilution processes, allowing you to adjust amounts accurately.
Solute
The solute is the substance that is dissolved within the liquid making up the solution. It is the component of a solution that is present in a smaller amount compared to the solvent (the dissolving liquid).
  • In this exercise, \(\mathrm{{H}_{2} \mathrm{SO}_{4}}\) is the solute being diluted.
  • The solute retains its chemical identity during dissolution.
Understanding the nature and role of the solute allows for precise adjustment of concentrations during experiments. Knowing how much solute is present helps in predicting the behavior and characteristics of the solution.
Solution
A solution is a homogeneous mixture composed of two or more substances. It consists of a solute that is dissolved in a solvent.
  • The solvent is typically present in greater amounts and determines the physical state of the solution.
  • In this case study, water acts as the solvent, diluting the concentrated sulfuric acid solution.
The properties of a solution, such as its concentration, are influenced by how much solute is dissolved at a particular temperature and pressure. Solutions are fundamental in chemistry, and understanding them is critical for performing laboratory work and chemical reactions.
Dilution Formula
The dilution formula \(C_1V_1 = C_2V_2\) is a straightforward equation used to calculate the concentration of a diluted solution.
  • \(C_1\) is the initial concentration, and \(V_1\) is the initial volume of the concentrated solution.
  • \(C_2\) represents the resulting concentration after dilution, while \(V_2\) is the final total volume of the diluted solution.
For the given problem involving \(\mathrm{{H}_{2} \mathrm{SO}_{4}}\), we set \(C_1 = 18.0 \, \mathrm{M}\), \(V_1 = 100.0 \, \mathrm{mL}\), and \(V_2 = 500.0 \, \mathrm{mL}\). Substituting these into the formula and solving for \(C_2\) yields the concentration of the new solution as 3.6 M. This demonstrates how dilution allows for precise adjustments in solution concentrations.

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Most popular questions from this chapter

Calculate the volume of \(0.0985 M\) sulfuric acid \(\left(\mathrm{H}_{2} \mathrm{SO}_{4}\right)\) that would be needed to react with \(10.89 \mathrm{~mL}\) of a \(0.01043 M\) aqueous ammonia \(\left(\mathrm{NH}_{3}\right)\) solution. $$ \mathrm{H}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{NH}_{3}(a q) \longrightarrow\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}(a q) $$

A 3.500 -gram sample of an oxide of manganese contains 1.288 grams of oxygen. What is the empirical formula of the compound?

Calculate the number of chlorine atoms in 0.756 gram of \(\mathrm{K}_{2} \mathrm{Pt} \mathrm{Cl}_{6}\)

What is the average mass in amu of one methane molecule? What is the mass in grams of one mole of methane?

Calculate the number of water molecules that can be prepared from \(500 \mathrm{H}_{2}\) molecules and \(500 \mathrm{O}_{2}\) molecules. $$ 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) $$ What would happen to the potential yield of water molecules if the amount of \(\mathrm{O}_{2}\) were doubled? What if the amount of \(\mathrm{H}_{2}\) were doubled?
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