Dilution and Concentration Calculations

The video tutorial covers solving dilution problems using the formula m1v1 = m2v2. It provides step-by-step solutions to three different problems involving sodium chloride, HCl, and NaOH. The tutorial emphasizes understanding the relationship between initial and final concentrations and volumes, and how to apply the formula to find unknown values. It also highlights the importance of checking the reasonableness of the results.

What is the initial volume of the sodium chloride solution in the first problem?

• 0.450 liters

• 1.800 liters

• 5.000 liters

• 0.850 liters

What is the final volume of the sodium chloride solution after dilution?

• 1.800 liters

• 5.000 liters

• 0.850 liters

• 0.450 liters

What is the concentration of the diluted sodium chloride solution?

• 5.00 M

• 2.36 M

• 0.85 M

• 1.80 M

What is the concentration of the stock solution of HCl in the second problem?

• 0.12 M

• 0.45 M

• 1.59 M

• 5.00 M

What volume of 0.12 M HCl can be prepared from 11 milliliters of 0.45 M HCl?

• 0.314 liters

• 11 milliliters

• 41 milliliters

• 5 liters

What is the concentration of the desired diluted HCl solution?

• 5.00 M

• 0.45 M

• 0.12 M

• 1.59 M

What is the concentration of the stock solution of NaOH in the third problem?

• 1.59 M

• 5.00 M

• 0.45 M

• 0.12 M

What volume of 1.59 M NaOH is required to prepare 5 liters of 0.1 M NaOH?

• 0.850 liters

• 5.000 liters

• 0.314 liters

• 1.800 liters

What is the final volume of the diluted NaOH solution?

• 1.800 liters

• 0.314 liters

• 5.000 liters

• 0.850 liters

What is the concentration of the desired diluted NaOH solution?

• 5.00 M

• 1.59 M

• 0.45 M

• 0.1 M