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Solutions Review. Play as a slideshow!. Calculate the molarity of a 0.175 L sugar solution that was prepared with 0.15 moles of sugar ? M = mol /L M = 0.15 mol /0.175 L M = 0.86 M. Solutions are ________________ mixtures made up of very small particles. Homogeneous. - PowerPoint PPT Presentation

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Page 1: Solutions Review

Solutions Review

Play as a slideshow!

Page 2: Solutions Review

Calculate the molarity of a 0.175 L sugar solution that was prepared with 0.15 moles of sugar?• M = mol/L• M = 0.15 mol/0.175 L• M = 0.86 M

Page 3: Solutions Review

Solutions are ________________ mixtures made up of very small particles.• Homogeneous

Page 4: Solutions Review

Determine the volume of 0.235 M sugar solution that can be prepared with 0.470 moles of sugar.• M = mol/L• 0.235 M = 0.470 mol/ x L ***cross-multiply• 0.235 x = 0.470• X = 2.00 L

Page 5: Solutions Review

Explain why solutions are classified as mixtures instead of compounds.• The two parts, solute and solvent, are only

physically combined, not chemically combined. Also, you can easily separate them with a physical change, such as evaporating off the solvent. Finally, they can combine in different ratios, they do not have a fixed ratio like compounds (example, water is always 2 H and one O, but salt water can have varying percent of salt in it).

Page 6: Solutions Review

A salt solution is to be added to a marine aquarium. Calculate the molarity of a salt solution that is prepared by adding water to 18.65 g of NaCl to give a final volume of 250.0 ml.• M = mol/L• 18.65 g x 1 mol/ 58.44 g = 0.3191 mol• M = 0.3191 mol/0.2500 L • M = 1.277 M

Page 7: Solutions Review

Pure gold is 24 carat. 14-carat gold contains 14 parts gold and 10 parts other metals. 14-carat gold is said to be a(n) ___________, which is a type of solution. An example of a gaseous solution is ________________, which is made up mostly of ______________ and nitrogen when dry. The most common solutions are ______________ solutions.

alloy

airoxygen

aqueous

Page 8: Solutions Review

Calculate the volume of a 3.15 M NaOH (aq) solution that should be used to prepare 250. ml of 0.150 M NaOH (aq).• M1V1 = M2V2

• 3.15 M x V1 = 0.150 M x 250. mL

• V1 = 11.9 mL

Page 9: Solutions Review

Because of the _______________ _______________, you can see the light beams from car headlights in a fog.• Tyndall Effect

Page 10: Solutions Review

You have 3.0 L of 3.0 M HCl. What volume of 2.0 M HCl can you make?• M1V1 = M2V2

• 3.0 M x 3.0 L = 2.0 M x V2

• V2 = 4.5 L

Page 11: Solutions Review

3.5 L of solvent was added to 2.0 L of a 0.88 M solution. What is the new molarity of the solution?• M1V1 = M2V2

• 0.88 M x 2.0 L = M2 x 5.5 L

• M2 = 0.32 M

Page 12: Solutions Review

Multiple choice: To increase the rate of solution of a solid in water,• a. increase the pressure over the water.• b. decrease the pressure over the water.• c. crush the particles of the solid.• d. chill the water.

Page 13: Solutions Review

You need 450 mL of 0.15 M NaOH. All you have available is a 2.0 M stock solution of NaOH. What volume of solvent will you add to the volume required of the stock solution to make your dilution?• M1V1 = M2V2

• 0.15 M x 450 mL = 2.0 M x V2

• V2 = 33.75 mL stock solution• 450 – 33.75 = 416.25 mL solvent

Page 14: Solutions Review

• If 15 grams of iodine are dissolved in 1000 mL of alcohol, the alcohol is the (solute, solvent) and the solution is said to be a(n) _______________.

• A substance that dissolves other materials is a (solute, solvent). The substance being dissolved is a (solute, solvent).

tincture

Page 15: Solutions Review

You have 6.0 L of 5.0 M NaCl stock solution. You take 800 mL of that stock solution, and add 500 mL of solvent.-How many moles of NaCl would be present in the new solution?-What is the molarity of new solution?• M = mol/L• 5.0 M = x mol/ 0.800 L• X = 4.0 mol• M1V1 = M2V2

• 5.0 M x 800 mL = M2 x 1300 mL

• M2 = 3.08 M

Page 16: Solutions Review

• In (solutions, suspensions) the substances separate after standing a while. The substances (can also, can not) be separated by filtration.

Page 17: Solutions Review

Find the volume of a 0.75 M solution if it contains 39 grams of KOH. • 39 g x 1 mol/56.1 g = 0.695 mol• M = mol/L• 0.75 M = 0.695 mol/x L ***cross-multiply***• X = 0.93 L

Page 18: Solutions Review

• Smoke is an example of a _____________ of solid dirt and dust particles in air.

• colloid

Page 19: Solutions Review

How many grams of hydrochloric acid (HCl) are present in 3.0 L of a 0.750 M solution? • M = mol/L• 0.750 M = x mol/ 3.0 L• X = 2.25 mol• 2.25 mol x 36.45 g/ 1 mol = 82.0 g

Page 20: Solutions Review

• A(n) ________________’s particles are between those of a solution and a suspension.

• colloid

Page 21: Solutions Review

110.0 mL of 3.00 M sulfuric acid has 25.0 mL of water added to it. What is the resulting concentration of the solution ?• M1V1 = M2V2

• 3.00 M x 110.0 mL = M2 x 135.0 mL

• M2 = 2.44 M

Page 22: Solutions Review

• How does a solution behave differently from a suspension when a beam of light is shined through it?

• A beam of light will go right through a solution and you will not see it. A suspension exhibits the Tyndall Effect where you will be able to see the beam of light.

Page 23: Solutions Review

Acetic acid is purchased as a solution in 17.0 M concentrations. Explain how you would prepare 500.0 mL of a 5.00 M solution.• M1V1 = M2V2

• 17 M x V1 = 5.00 M x 500.0 mL

• V1 = 147.1 mL• Put 147.1 mL into a 500 mL volumetric flask.

Then add water until the solution is 500 mL total (to the line).

Page 24: Solutions Review

• What is the solute in a brass alloy containing 75% copper and 25% zinc?

• Zinc (it has a smaller percentage)

Page 25: Solutions Review

• Substances that conduct electricity when dissolved are said to be ______________, while substances that do NOT conduct electricity when dissolved are said to be __________________.

electrolytes

Non-electrolytes

Page 26: Solutions Review