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Ch.13 - Properties of Solutions
Chapter 13, Problem 44c

What is the molarity of each of the following solutions: (c) 35.0 mL of 9.00 M H2SO4 diluted to 0.500 L?

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1
Start by identifying the initial concentration and volume of the solution. Here, the initial concentration (C1) is 9.00 M and the initial volume (V1) is 35.0 mL.
Convert the initial volume from milliliters to liters to ensure consistency in units. Use the conversion: 1 mL = 0.001 L. Therefore, V1 = 35.0 mL * 0.001 L/mL.
Recognize that the problem involves dilution, which can be solved using the dilution equation: C1 * V1 = C2 * V2, where C2 is the final concentration and V2 is the final volume.
Substitute the known values into the dilution equation. You have C1 = 9.00 M, V1 in liters from the previous step, and V2 = 0.500 L. Solve for C2, the final concentration.
Rearrange the equation to solve for C2: C2 = (C1 * V1) / V2. Substitute the values and calculate to find the molarity of the diluted solution.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Molarity

Molarity (M) is a measure of concentration defined as the number of moles of solute per liter of solution. It is expressed in moles per liter (mol/L) and is crucial for understanding how much solute is present in a given volume of solution. Molarity is commonly used in chemistry to prepare solutions and perform stoichiometric calculations.
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Dilution

Dilution is the process of reducing the concentration of a solute in a solution, typically by adding more solvent. The dilution equation, M1V1 = M2V2, relates the initial and final molarities (M1 and M2) and volumes (V1 and V2) of the solution. This concept is essential for calculating the new concentration after a solution has been diluted.
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Volume Conversion

Volume conversion is the process of changing the volume of a solution from one unit to another, such as milliliters (mL) to liters (L). Since molarity is expressed in terms of liters, it is important to convert volumes appropriately when performing calculations. Understanding how to convert between these units is vital for accurate concentration calculations.
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