Molarity and Dilution Concepts

Mr. Craig explains how to calculate the concentration of barium ions in a stock solution and a diluted solution. The process involves preparing a stock solution by dissolving barium metal in nitric acid, calculating its concentration, and then diluting a portion of it to find the new concentration. Key concepts include understanding moles, molarity, and the importance of final volume in concentration calculations.

What is the initial mass of barium metal used in the stock solution preparation?

• 23 grams

• 25 grams

• 20 grams

• 30 grams

What is the final volume of the stock solution prepared?

• 1 liter

• 2 liters

• 3 liters

• 4 liters

How many moles of barium ions are present in the stock solution?

• 0.1500 moles

• 0.2000 moles

• 0.1800 moles

• 0.1675 moles

What is the molarity of the stock solution?

• 0.5000 M

• 1.0000 M

• 0.7500 M

• 0.8374 M

How much of the stock solution is used to prepare the dilute solution?

• 150 mL

• 125 mL

• 175 mL

• 100 mL

What is the final volume of the dilute solution?

• 500 mL

• 550 mL

• 450 mL

• 400 mL

What is the molarity of the new dilute solution?

• 0.2500 M

• 0.3000 M

• 0.2093 M

• 0.1500 M

What remains constant when water is added to create a dilute solution?

• Temperature

• Concentration

• Moles

• Volume

What should you do if the dilute solution's molarity is higher than the stock solution?

• Use more barium

• Add more water

• Check calculations

• Increase temperature

What is the key concept to remember about moles in a dilute solution?

• Moles remain the same

• Moles decrease

• Moles increase

• Moles depend on temperature