Molarity and Dilution Problem Solving

This video tutorial covers the concept of molarity, including how to calculate it from grams and volume, and how to handle dilution and concentration changes. It also explains acid-base titration and stoichiometry for reaction calculations, and concludes with methods for calculating concentrations in mixtures and solutions.

What does the dilution equation M1V1 = M2V2 represent?

• The change in molarity due to temperature

• The type of solute used in the solution

• The volume of the solution after dilution

• The concentration of solute remains constant after dilution

How can you find the number of moles from molarity and volume?

• Add molarity to volume

• Subtract volume from molarity

• Divide molarity by volume

• Multiply molarity by volume

What is required to calculate the mass of a substance from its molarity and volume?

• The density of the substance

• The electrical conductivity of the solution

• The molar mass of the substance

• The boiling point of the substance

Which method can increase the concentration of a solution?

• Adding more solvent

• Decreasing the volume by evaporation

• Diluting with more water

• Increasing the temperature

In a reaction where zinc displaces hydrogen from hydrochloric acid, what is the product formed with chlorine?

• Chlorine gas

• Zinc chloride

• Zinc hydroxide

• Hydrogen gas

What is the molar mass of aluminum used for converting grams to moles?

• 24.31 g/mol

• 39.10 g/mol

• 22.99 g/mol

• 26.98 g/mol

What is the mole ratio of aluminum to copper chloride in the discussed reaction?

• 2:3

• 1:1

• 3:2

• 1:2

How is the solubility of lead chloride described in the reaction with sodium?

• Not mentioned

• Partially soluble

• Soluble

• Insoluble

Which reactant is the limiting reactant when aluminum reacts with sulfuric acid?

• Aluminum

• Sulfuric acid

• Hydrogen gas

• None of the above

What is the theoretical yield of hydrogen gas when 10 g of aluminum reacts with sulfuric acid?

• 1.25 g

• 0.50 g

• 2.00 g

• 1.12 g