Solutions Packet

1 Solutions











Chemistry: Form WS8.1.1A Name ______________________________ 

SOLUTIONS Date _________________ Period _____ 

What’s in the Water? 

A solution is a type of mixture. It consists of two or more kinds of matter 

each of which retains its own properties. But it is homogeneous. It appears 

to be only one substance. This is what distinguishes solutions from 

mechanical mixtures. The substance that is dissolved is the solute. The 

solvent is the substance that dissolves the solute. It is the continuous phase. 

For example, salt dissolved in water appears to be a liquid. 

Different solutes dissolve best in different solvents. In order for a 

solvent to dissolve a solute, it must exert forces of attraction on the solute. 

Polar solvents such as water dissolve polar and ionic solutes well because 

they exert mutual attractions that cause their particles to intermingle. Of 

course, not all ionic substances dissolve equally well in water. (See Table 

F - Table of Solubilities in Water) Nonpolar solvents do NOT dissolve 

polar and ionic substances because there is no attraction between them. For 

example, oil and water do NOT mix. Nonpolar substances such as fat 

dissolve in nonpolar solvents such as benzene because the forces of 

attraction are too weak to prevent the particles from freely intermingling. 

Answer the questions below based on your reading and on your 

knowledge of chemistry. 

1. Water is mixed with sugar, resulting in a transparent, colorless liquid. 

a. What evidence will there be that this is a mixture rather than a new compound? 

b. What evidence shows it is a solution rather than a mechanical mixture? 

c. Which is the solute, and which is the solvent? How do you know? 

2. Why does table salt dissolve in water, while oil and water don’t mix? 

© Evan P. Silberstein, 2003 

Water discusses its tastes. 

3. Based on Table F indicate which of the following compounds is water soluble and which is insoluble? 

a. Li 2CO 3 _______________ 

b. Fe(OH) 3 _______________ 

c. CaCrO 4 _______________ 

d. BaS _______________ 

e. (NH 4) 3PO 4 _______________ 

f. Al(ClO 3) 3 _______________ 

g. PbSO 4 _______________ 

h. NaOH _______________ 

i. CuSO 4 _______________ 

j. KNO 3 _______________ 

k. AgCl _______________ 

l. Ba(HCO 3) 2 _______________ 




Solubility Curves 

The solubility of solid solutes generally increases as temperature increases, while 

the solubility of gaseous solutes generally decreases as temperature increases. A 

solution that holds as much solute as can dissolve at a given temperature is 

saturated. A solution that can dissolve more solute at a given temperature is 

unsaturated. A solution that holds more solute than can dissolve at a given 

temperature is supersaturated. The amount of solute that is needed to form a 

saturated solution at various temperatures can be graphed. This is what is shown 

in Table G. The values in Table G are based on solute dissolved in 100 g of water. 

Since water has a density of 1 g/mL, the graph can be considered to be based on 

100 mL of water. A 200 mL sample of water would be able to dissolve twice as 

much at each temperature. 

Answer the questions below by referring to Table G. 

1. The compound which is the most soluble at 20°C is . 

2. The compound which is the least soluble at 10°C is . 

3. The compound which is the least soluble at 80°C is . 

4. The number of grams of potassium nitrate needed to saturate 100 mL of water 

at 70°C is . 

5. The formulas of the compounds which vary inversely with the temperature are , 

and . 

6. One hundred mL of a sodium nitrate solution is saturated at 10°C. How many additional grams are needed to saturate 

the solution at 50°C? 

7. One hundred mL of a saturate KCl solution at 80°C will precipitate 10 grams of salt when cooled to what temperature? 

8. The two salts that have the same degree of solubility at 70°C are and . 

9. The salt with a solubility is least affected by a change in temperature is . 

10. The salt that has the greatest increase in solubility in the temperature range between 30°C and 50°C is . 

11. The number of grams of sodium nitrate that must be added to 50 mL of water to produce a saturated solution at 50°C is 

12. A saturated solution of potassium chlorate is made at 10°C by dissolving the correct mass of salt in 100 mL of water. 

When the solution is heated to 90°C, how many grams must be added to saturate the solution? 

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13 Solutions 

.

Chemistry: Form WS8.1.2A Solubility Curves 

SOLUTIONS Page 2 

13. At what temperature do saturated solutions of sodium chloride and potassium chloride contain the same mass of solute 

per 100 mL of water? 

14. A saturated solution of potassium nitrate is prepared at 60°C using 200 mL of water. If the solution is cooled to 30°C, 

how many grams will precipitate out of the solution? 

15. How many more grams of ammonia can be dissolved in 100 mL of water at 10°C than at 90°C? 

16. A saturated solution of sodium nitrate in 100 mL of water at 40°C is heated to 50°C. The rate of increase in solubility 

grams per degree is . 

17. Thirty grams of KCl is dissolved in 100 mL of water at 45°C. The number of additional grams of KCl that would be 

needed to make the solution saturated at 80°C is . 





Dissolving Solids and Gases 

A factory releases clean, warm water into a stream. The stream becomes 

severely polluted as a result. How does this happen? Fish living in the water 

depend on dissolved oxygen in order to breathe. Like other gases, oxygen 

molecules tend to spread out. In order to dissolve them, it is necessary to 

confine them. Heat speeds the molecules up and makes them spread out 

more–exactly the opposite of what is needed to dissolve them. As a result, 

heat drives the oxygen out of the water, causing the fish to die. The dead 

fish begin to decay. Growing decay bacteria deplete the water of oxygen 

even further. In this way, clean warm water can pollute a stream. The 

process of dissolving gases is opposite to the process of dissolving solids 

because of the differences between gases and solids. 

Answer the questions below based on your reading above and on your knowledge of chemistry. 

1. A warm can of soda is dropped and bounces down a flight of stairs. When it is opened, carbon dioxide gas 

coming out of solution causes it to spray all over. Explain the affect of each of the following: 

a. The fact that the soda was warm. 

b. The fact that the soda was dropped and bounced down a flight of stairs. 

c. The fact that the can was opened. 

2. When a gas dissolves, the particles need to be confined. What do the particles of a solid need to do in order to 

dissolve? 

3. Sugar is added to a hot cup of coffee and stirred. The sugar dissolves. Explain the affect of each of the 

following: 

a. The fact that the coffee was hot. 

b. The fact that the coffee was stirred. 

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Chemistry: Form WS8.1.3A Dissolving Solids and Gases 


4. Which dissolves faster, a teaspoon of sugar or a sugar cube? Why? 

5. A solid is added to water and stirred. Some of it dissolves, but not all. What happens to the rate at which the 

solid is dissolving between when it was first added and when it stopped dissolving? Explain. (HINT: 

Equilibrium!) 

6. The table below lists four factors that may effect the rate at which solids and gases dissolve. Fill in the table by 

indicating if the rate of dissolving increases, decreases, or is not effected. Then explain why. 

Crushing 

Stirring 

Factor 

Increasing the 

amount of dissolved 

solute 

Increasing 

Temperature 

Affect on Rate of Solution for: 

Solid Solutes Gaseous Solutes 





Finding Concentration 

The directions on a can of condensed soup say to mix the can of soup with one can 

of water. What would happen to the flavor if it were mixed with two or three cans 

of water instead? When two substances are mixed, the amount of one compared to 

the amount of the other is known as the concentration. Adding extra water makes 

the concentration of the soup lower than what is called for in the recipe–and it tastes 

it! There are several ways of measuring concentration–mass per unit volume, 

percentage by mass, percentage by volume, and parts per million (ppm). See the 

examples below: 

Massof solute( g) 

Concentration = 

Volumeof Solvent or Solution( mL) 

Sample Problem 

What is the concentration of a solution prepared by 

dissolving 25 g of KNO 3 in 100. mL of water? 

25g 

Concentration = = 025 . 

100. 

ml 

mass( solute) 

percent mass = × 

mass( solution) 

100% 

g 

mL 


What is the percent by mass of a solution containing 12.3 

g of caffeine dissolved in 100.0 g of water? 

Step 1: Find the mass of the solution 

100.0 g + 12.3 g = 112.3 g 

Step 2: Divide the mass of the solute by the mass of the 

solution and multiply by 100 % 

12. 3g 

percent mass = × 100% = 110% . 

112. 3g 

Continue L 

mass( solute) 

ppm = × 1, 000, 000ppm 

mass( solution) 


About 0.0047 g of ammonia are dissolved in 20.0 g of 

water. Express this in parts per million. 

Step 1: Find the mass of the solution 

20.0 g + 0.0047 g = 20.0047 g 

Step 2: Divide the mass of the solute by the mass of the 

solution and multiply by 1,000,000 ppm. 

0. 0047g 

ppm = × 1, 000, 000ppm = 235ppm 

20. 0047g 

volume ( solute) 

percent volume = × 

volume ( solution) 

100% 


What is the percent by volume of a solution containing 

18.2 mL of glycerine (C 3H 6O 3) dissolved in 85.0 mL of 

water? 

Step 1: Find the volume of the solution. 

18.2 mL + 85.0 mL = 103.2 mL 

Step 2: Divide the volume of the solute by the volume of 

the solution and multiply by 100% 

18. 2mL 

percent volume = × 100% = 17. 6% 

1032 . mL 


Chemistry: Form WS8.2.1A Finding Concentration 


Answer the questions below based on the sample problems. 

1. What is the concentration of 45 mL of a solution 

containing 9.0 g of KClO 3? 

2. A solution is prepared by mixing 20.0 g of NaNO 3 with 

100. mL of water. What is the percentage mass of the 

solution? (Assume density of water is 1 g / mL) 

3. A 250. mL sample of air at STP contains approximately 

52.5 mL of O 2(g). What is the percentage of oxygen in 

air? 

4. A polar solvent is prepared by mixing 27.5 mL of 

propanone with 222.5 mL of water. What is the 

percentage by volume of propanone in the mixture? 

5. How many parts per million of sulfur dioxide are there 

in a solution containing 0.065 g of sulfur dioxide in 

5,000 mL of water? (Assume density of water is 1 g / mL) 


6. If 19 mL of alcohol are dissolved in 31 mL of water, 

what is the percentage by volume of alcohol? 

7. If 0.002 g of PbCl 2 are dissolved in 2.0 L of water, how 

many parts per million are dissolved? (Assume density 

of water is 1 g / mL) 

8. If 15 g of KNO 3 are dissolved in 235 g of water, what is 

the percentage of solute by mass? 

9. What is the percentage by mass of a solution prepared 

with 34 g of KI and 126 g of water? 

10. What is the concentration of a solution made with 

0.056 g of CO 2(g) and 200 mL of water? 




Molarity 

One of the most useful measures of concentration in chemistry is molarity (M). Molarity is the 

number of moles of solute per liter of solution. A two molar (2 M) solution contains two moles of 

solute per liter of solution. 

moles( solute) 

M = 

L( solution) 

Recall that the number of moles is determined by dividing the number of grams by the gram formula 

mass (GFM). There are a number of formulas for calculation that come from these relationships: 

• 

g 

• moles = M × L • g = M × GFM × L 

M = 

GFM × L 

Below are some sample problems that show how to apply these formulas. 

Sample Problem 1 

Find the molarity of 100. mL of a solution that contains 

0.25 moles of dissolved solute. 

Step 1: Convert all volumes to liters 

0001 . L 

100. 

mL × = 0100 . L 

1mL 

Step 2: Substitute values into the definitional equation 

M mol 025 . mol 

= = = 25 . M 

L 0100 . L 


How many moles of solute are dissolved in 250. mL of a 

3.0 M solution? 


0. 001L 

250. 

mL × = 0250 . L 

1mL 

Step 2: Substitute values into the correct equation 

mol = M × L = 30 . 0. 250L = 0. 75mol 

mol ( L)( 

) 

A two molar solution 


Find the molarity of 500. mL of a solution that contains 

4.9 g of dissolved sulfuric acid (H 2SO 4). 

Step 1: Find the GFM 

H = 1 × 2 = 2 

S = 32 × 1 = 32 

O = 16 × 4 = 64 

98 


0001 . L 

500. 

mL × = 0500 . L 

1mL 


g 

M = = 

GFM × L 98 

49 . g 

0500 . L 

= 010 . M 

g ( mol)( 

) 


How many grams of sodium carbonate(Na 2CO 3) are 

needed to prepare 250 mL of a 0.10 M solution? 


Na = 23 × 2 = 46 

C = 12 × 1 = 12 

O = 16 × 3 = 48 

106 


0. 001L 

250. 

mL × = 0250 . L 

1mL 


g = M × L× GFM = 010 . 106 0. 250L = 2. 7g 

mol g ( L)( 

mol)( 

) 


Chemistry: Form WS8.2.2A Molarity 


Answer the questions below based on the reading and the sample problems on the previous page. 

1. Determine the molarity of 500. mL of a solution with 

0.35 mol of dissolved solute. 

2. A 200. mL sample of a solution contains 4.0 g of NaOH. 

What is its molarity? 

3. How many grams of KNO 3 are needed to prepare 25 mL 

of a 2.0 M solution? 

4. How many moles of MgSO 4 are contained in 50. mL of 

a 3.0 M solution? 

5. How many grams of CaCl 2 are dissolved in 80.0 mL of 

a 0.75 M solution? 


6. What is the molarity of 300 mL of a solution that 

contains 0.60 mol of dissolved ammonia? 

7. What is the molarity of 5.0 L of a solution containing 

200. g of dissolved CaCO 3? 

8. How many grams of NaCl are needed to prepare 500. 

mL of a 0.400 M solution? 

9. How many moles of solute are contained in 3.0 L of a 

1.5 M solution? 

10. What is the molarity of 750 mL of a solution that 

contains 40.0 g of dissolved CuSO 4? 




Understanding Colligative Properties 

After a winter storm, people spread salt on the walks to help melt the ice. 

Salt reduces the freezing point of water. Actually, any soluble solute 

reduces the freezing point of water by interfering with crystallization. In 

this way, antifreeze keeps the water from freezing in an automobile 

radiator. This phenomenon is called freezing point depression. Antifreeze 

is left in the radiator during the summer. It also prevents the radiator from 

boiling over by raising the boiling point. Dissolved solute reduces the 

vapor pressure, raising the boiling point. This is called boiling point 

elevation. 

The amount the freezing point 

is depressed or the boiling 

point is raised depends on the 

concentration of dissolved 

solute. The higher the Dad misinterprets freezing point depression. 

concentration of dissolved 

solute is, the greater the effect on the boiling point or the freezing point is. Only the 

concentration of the particles of dissolved solute is important. The nature of the 

solute is not. A mole of dissolved sugar has exactly the same effect on the freezing 

point and boiling point of 1,000 g of water as a mole of antifreeze because it 

contains the same number of particles. Ionic compounds dissociate producing 

more particles per mole. One mole of dissolved sodium chloride, for example, produces one mole of aqueous sodium ions and 

one mole of aqueous chloride ions for a total of two moles [NaCl(s) ÿ Na + (aq) + Cl – (aq)]. One mole of dissolved sodium 

chloride, therefore, has twice the effect on the boiling and freezing points of 1,000 g of water as one mole of dissolved sugar. 

It is not the nature of the solute that matters, but only the concentration of dissolved particles that determines how large the 

change in freezing point or boiling point will be. Properties of a solution, such as this, which are dependent only on the number 

of particles in solution, and not on their nature are called colligative properties. 

Answer the questions below based on your reading and on your knowledge of chemistry. 

1. Why are boiling point elevation and freezing point depression considered colligative properties? 

2. Why is salt put on icy roads and sidewalks in the winter? 

3. How will the boiling points of pure water and sea water compare? Why? 

Continue L 




Molality 

There are two basic ways to prepare solutions quantitatively. Keep in mind that 

solute takes up space. Even when the solute is completely dissolved, it affects the 

volume of the solution. Preparing solutions based on measuring the amount of 

solute per volume of solution is useful, because it is possible to measure out a 

sample of the solution and figure out how much solute you have. This is the type 

of measure that molarity is. Understanding how solute affects the solution, 

however, requires a different type of measure. To compare the solubility of 

different substances, it is necessary to know how much solvent is used compared 

to solute. To understand how dissolved solute affects the freezing point or boiling 

point of a liquid, it is necessary to know how much liquid solvent you have for a 

given amount of solute. Molality is very similar to molarity, except that it compares 

the moles of solute to kilograms of solvent instead of liters of solution. It is 

abbreviated by lower case m. 

mol( solute) 

m = 

kg( solvent) 


Find the molality of a solution that contains 0.45 moles of solute 

dissolved in 300. g of water . 

Step 1: Convert the amount of solvent to kilograms 

1kg 

300. 

g × = 0. 300kg 

1000g 

Step 2: Substitute values into the definitional equation 

045 . mol 

m = = 15 . m 

0300 . kg 

Answer the questions below. 

1. Find the molality of a solution that 

contains 225 g of Ca(NO 3) 2 

dissolved in 400. g of water. 


contains 0.663 mol of solute 

dissolved in 300. g of water. 

Competition among molar and molal 

solutions 


Find the molality of a solution that contains 25.57 g of sodium 

chloride dissolved in 250. g of water. 


Na = 22.99 × 1 = 22.99 

Cl = 35.45 × 1 = 35.45 

58.44 

Step 2: Convert the mass of solute to moles. 

⎛ 1mol 

⎞ 

( 2557 . g) 

⎜ ⎟ = 0. 4375mol 

⎝ 58. 44g 

⎠ 

Step 3: Convert the amount of solvent to kilograms. 

1kg 

250. 

g × = 0. 250kg 

1000g 

Step 4: Substitute values into the definitional equation. 

0. 4375mol 

m = = 175 . m 

0250 . kg 


contains 1.25 kg of KBr dissolved 

in 2.45 kg of water. 



Chemistry: Form WS8.3.1A Understanding Colligative Properties 


Solve the following boiling point elevation problems and the freezing point depression problems as shown in the sample 

problems below. [NOTE: At standard pressure, 1 mol of dissolved particles will elevate the boiling point of 1,000 g of 

water by 0.52EC and will depress the freezing point of 1,000 g of water by 1.86EC] 


Find the boiling point of a solution containing 1,000 g 

of water and 2 mol of dissolved MgF 2. 

Step 1: Determine the number of moles of solute 

particles 

2MgF 2(s) ÿ 2Mg 2+ (aq) + 4F – (aq) mol =6 

Step 2: Multiply the boiling point elevation per mole by 

the number of moles of solute to find the boiling 

point elevation 

BPE = 0.52 EC /mol × 3 mol = 3.12EC 

Step 3: Add the boiling point elevation to 100EC 

BP = 100EC + 3.12EC = 10 3.12EC 


Find the freezing point of a solution containing 1,000 g of 

water and 30 g of dissolved antifreeze (C 2H 4O 2). 

Step 1: Determine the number of moles of solute particles 

C = 12 × 2 = 24 g 30g 

H = 1 × 4 = 4 

mol = = = 05 . mol 

g GFM 60 mol 

O = 16 × 2 = 32 

60 

Step 2: Multiply the freezing point depression per mole by the 

number of moles of solute to find the freezing point 

depression 

FPD = 1.86EC /mol × 0.5 mol = 0.93EC 

Step 3: Subtract the freezing point depression from 0EC 

FP = 0EC – 0.93EC = – 0.93EC 

4. One mole of dissolved particles elevates the boiling point of 1,000 g of water by 0.52EC. At standard pressure, what will 

the boiling point of a solution be if it contains 1,000 g of water and: 

a. 1 mol of antifreeze (C 2H 4O 2)? _______________ 

b. 1 mol of salt (NaCl)? _______________ 

c. 1 mol of ethanol (C 2H 5OH)? _______________ 

d. 2 mol of glycerol (C 3H 6O 3)? _______________ 

e. 2 mol of CaCl 2(aq)? _______________ 

f. 5 mol of sucrose (C 12H 22O 11)? _______________ 

g. 1 mol of KNO 3(aq)? _______________ 

h. 3 mol of Ba(NO 3) 2(aq)? _______________ 

i. 40 g of NaOH(aq)? _______________ 

j. 270 g of glucose (C 6H 12O 6)? _______________ 

5. One mole of dissolved particles depresses the freezing point of 1,000 g of water by1.86EC. At standard pressure, what 

will the freezing point of a solution be if it contains 1,000 g of water and: 

a. 1 mol of glucose (C 6H 12O 6)? _______________ 

b. 1 mol of BaCl 2(aq)? _______________ 

c. 2 mol of methanol (CH 3OH)? _______________ 

d. 3 mol of glycerol (C 3H 6O 3)? _______________ 

e. 2 mol of CuSO 4(aq)? _______________ 


f. 4 mol of sucrose (C 12H 22O 11)? _______________ 

g. 3 mol of KNO 3(aq)? _______________ 

h. 2 mol of salt (NaCl)? _______________ 

i. 150 g of KHCO 3(aq)? _______________ 

j. 180 g of glucose (C 6H 12O 6)? _______________ 


Name: _______________________________ 

Solubility Practice Sheet 

1. If 50g of water is saturated with Potassium chlorate at 23°C is evaporated, how 

many grams of the salt will be left? 

2. What is the smallest mass of water needed to dissolve 23g of ammonium chloride 

at 40°C? 

3. At what temperature do potassium chlorate and potassium chloride have the same 

solubility in water? 

4. If you have 30g of potassium chloride already dissolved in 100g of water, how 

many more grams are needed to make a saturated solution at 80°C? 

5. Which substances decrease in solubility as the temperature increases? Why? 

6. Which substance’s solubility is least affected by temperature changes? 

7. How much potassium nitrate will dissolve in 300g of water at 40°C? 

8. If 400g of water is mixed with 250g of ammonium chloride and is then cooled to 

25°C, how many grams will precipitate out of the solution? 


Regents Chemistry Review: Solutions 

Solution: a homogenous mixture of substances in the same physical state. Solutions contain atoms, ions or molecules of one substance 

spread uniformly throughout a second substance. Solutions will pass through filter paper and will not settle on standing. 

Solute: the substance that is being dissolved and the substance that is typically present in the smaller amount 

Solvent: the substance that dissolves the solute, and is typically present in the greater amount. 

Solubility: the amount of solute that will dissolve at a certain temperature ( Table G ) 

Table F tells if a compound is soluble or insoluble based on the ions that make it up. 

Factors that determine solubility: 

- Nature of solute and solvent: “like dissolves like” 

polar dissolves polar, nonpolar dissolves nonpolar 

- Temperature- Increase temperature, increase solubility of solids, 

Increase temperature, decrease the solubility of gases 

- Pressure- only affects gases; as Pressure increases, increase the 

solubility of gases in liquids. 

Reading Table G: Seven Salts, Three Gases! 

Unsaturated: below the curve; a point where the solution holds less 

solute than the maximum it can hold 

Saturated: on the curve: a solution that contains a maximum amount of solute 

that will dissolve at a specific temperature; any additional solute will 

not dissolve, it will simply settle to the bottom of the container. 

Supersaturated: above the curve: a solution that holds more solute than is present in a 

saturated solution at that temperature. These are unstable and temporary . 

Concentration of Solutions; use Table T to find the formulas 

Molarity: the number of moles of solute in 1 L of solution. (Concentration) 

**Solution must be in Liters! 

Percent by Mass: the mass of an ingredient divided by the total mass, expressed as a percent. 

Percent by Volume: the ratio of the volume of an ingredient divided by the total volume and expressed as a percent. 

Parts per million: a ratio between the mass of solute and the total mass of solution. The only difference between ppm and 

percent by mass is that you multiply by 1,000,000 instead of 100 

Preparing a solution of known concentration: Add solute to a small amount of solvent and then top off the solvent to make the given volume of 

the solution. First add sufficient grams to the flask, and then add distilled water until the solute is dissolved and thoroughly mixed. Finally, fill with 

distilled water to the mark on the neck of the flask, and again stir to make sure the solution is homogeneous. **Remember how we made Kool-aide** 

Colligative Properties : the boiling and the freezing point of water change when nonvolatile solutes are added. 

Covalent molecules do NOT separate in solution, so they do not lower the freezing point as much as ionic 

compounds. The more ions in the solution, the lower the freezing point, the higher the boiling point. 

Freezing point depression: when you add moles of particles, the freezing point gets lower 

3.0 M MgCl2 lowers the freezing point more than does 3.0 M NaCl because MgCl 2 ionizes into 3 moles of ions, as 

opposed to NaCl which only ionizes into 2 moles of ions. 3.0 M C 6H 12O 6 lowers the freezing point, but not as 

much as the ionic compounds because covalent molecules do not ionize at all. 

Boiling point elevation: when you add moles of particles, the boiling point gets higher 

3.0 M MgCl2 increases the boiling point more than does 3.0 M NaCl because MgCl 2 ionizes into 3 moles of ions, as 

opposed to NaCl which only ionizes into 2 moles of ions. 3.0 M C6H 12O 6 lowers the freezing point, but not as 

much as the ionic compounds because covalent molecules do not ionize at all.

Solutions Packet

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