Preparing and Diluting Solutions

This tutorial introduces quantitative analysis of reactions in aqueous solutions, focusing on molarity as a key concept. It explains how to prepare solutions, perform dimensional analysis, and calculate solute mass and solution volume. The video also covers dilution and stock solutions, providing examples of calculations using molarity and volume.

What is the primary focus when turning from qualitative to quantitative aspects of reactions in aqueous solutions?

• Identifying the pH level of solutions

• Measuring the temperature of reactions

• Describing the concentration of solutions

• Understanding the color change in reactions

Which unit is most practical for concentration when dealing with stoichiometry in aqueous solutions?

• Grams per liter

• Molarity

• Parts per million

• Percentage

What is the formula for calculating molarity?

• Grams of solute per milliliter of solution

• Grams of solute per liter of solution

• Moles of solute per milliliter of solution

• Moles of solute per liter of solution

What is the first step in preparing a solution?

• Measuring the desired amount of solute

• Mixing the solute with water

• Heating the solution

• Adding the solute to a volumetric flask

In the preparation of a solution, what is used to ensure the correct final volume?

• A beaker

• A volumetric flask

• A graduated cylinder

• A measuring cup

How many grams of potassium carbonate are needed to prepare 200 mL of a 2.5 M solution?

• 138.2 grams

• 69.1 grams

• 100 grams

• 34.5 grams

What is the purpose of using a stock solution?

• To prepare a solution from a dry reagent

• To create a more concentrated solution for later dilution

• To measure the pH of a solution

• To increase the concentration of a solution

What remains constant during the dilution process?

• Temperature of the solution

• Volume of the solution

• Molarity of the solution

• Moles of solute

Which formula is used for dilution calculations?

• D1V1 = D2V2

• P1V1 = P2V2

• C1V1 = C2V2

• M1V1 = M2V2

How many milliliters of a 2.5 M sodium hydroxide solution are needed to prepare 525 mL of a 0.15 M solution?

• 100 mL

• 31.5 mL

• 200 mL

• 52.5 mL