Classification of Classification of Matter and Solutions Matter and Solutions

Notes PartNotes Part 11

I. Classification of MatterI. Classification of MatterMatterMatter

Can it be physically separated?Can it be physically separated?YesYes NoNo

MixturesMixtures Pure SubstancesPure SubstancesIs the composition uniform?Is the composition uniform? Can it be Can it be decomposed by an decomposed by an

ordinary chemical reaction?ordinary chemical reaction?Yes Yes NoNo Yes Yes NoNo

HomogeneousHomogeneous HeterogeneousHeterogeneous CompoundsCompounds ElementsElements

MixturesMixtures MixturesMixtures (water, sodium(water, sodium (gold, oxygen,(gold, oxygen,

(Solutions)(Solutions) (Suspensions(Suspensions chloride, sucrose)chloride, sucrose) carbon)carbon)

(air, sugar water,(air, sugar water, or Colliods)or Colliods)salt water)salt water) (granite, wood, (granite, wood,

muddy water)muddy water)

MixturesMixtures: matter that can be : matter that can be physically separated into component physically separated into component parts. parts.

a. a. homogeneous mixturehomogeneous mixture –has –has uniform composition; also called a uniform composition; also called a solutionsolution

b. b. heterogeneous mixture heterogeneous mixture – does – does not have a uniform composition; not have a uniform composition; suspensions or colloidssuspensions or colloids

Pure SubstancesPure Substances: when component : when component parts of a mixture can no longer be parts of a mixture can no longer be physically separated into simpler physically separated into simpler substances. Pure substances are either substances. Pure substances are either compounds or elements.compounds or elements.

a. a. CompoundsCompounds – can be decomposed by – can be decomposed by a chemical change.a chemical change.

b. b. ElementsElements – cannot be decomposed by – cannot be decomposed by a chemical change.a chemical change.

II. Types of Mixtures (Solutions, II. Types of Mixtures (Solutions, Suspensions, and Colloids) Table 13-3 Suspensions, and Colloids) Table 13-3

page 398page 398

1. 1. Solution (homogeneous mixture)Solution (homogeneous mixture)- any - any substance (solid, liquid, gas) that is evenly substance (solid, liquid, gas) that is evenly dispersed throughout another substance. dispersed throughout another substance. page 398 (Not the same as a chemical page 398 (Not the same as a chemical reaction!!)reaction!!)

Ex: sugar water, salt water (do not scatter Ex: sugar water, salt water (do not scatter light)light)

Components of a SolutionComponents of a Solution 1. 1. Solute Solute – substance dissolved– substance dissolved 2. 2. SolventSolvent – substance that does the – substance that does the

dissolving dissolving (water is the universal solvent)(water is the universal solvent)

2. 2. Suspensions (heterogeneous Suspensions (heterogeneous mixtures)mixtures) – particles in a solvent – particles in a solvent are so large that they settle out are so large that they settle out unless the mixture is constantly unless the mixture is constantly stirred Ex: muddy water, vegetable stirred Ex: muddy water, vegetable soup, page 398 (may scatter light, soup, page 398 (may scatter light, but are transparent)but are transparent)

3. 3. Colloids (heterogeneous mixtures) Colloids (heterogeneous mixtures) – – particles are intermediate in size particles are intermediate in size between those in solutions and between those in solutions and suspensions. Example: After large soil suspensions. Example: After large soil particles settle out of muddy water the particles settle out of muddy water the water is often still cloudy because water is often still cloudy because colloidal particles remain dispersed in colloidal particles remain dispersed in the water. Ex: milk, mayonnaise , page the water. Ex: milk, mayonnaise , page 398 (do scatter light – Tyndall Effect) 398 (do scatter light – Tyndall Effect)

III. The Solution Process (Solvation)III. The Solution Process (Solvation)Solvation is the process by which a solute Solvation is the process by which a solute

dissolves in a solvent. dissolves in a solvent. MiscibleMiscible: when solutes and solvents are : when solutes and solvents are

soluble in each other (solvation occurs)soluble in each other (solvation occurs)ImmiscibleImmiscible: when solutes and solvents : when solutes and solvents

are not soluble in each other (solvation are not soluble in each other (solvation does not occur)does not occur)

Aqueous solutions – solvent is water.Aqueous solutions – solvent is water.

What happens when:What happens when:Ionic compound as the soluteIonic compound as the soluteand water is the solvent?and water is the solvent?Dissociation of ionic compound occurs Dissociation of ionic compound occurs

(ions separate). Water is then attracted (ions separate). Water is then attracted to the positive and negative ions. When to the positive and negative ions. When all molecules have been “surrounded” all molecules have been “surrounded” the molecule is called hydrated.the molecule is called hydrated.

MiscibleMiscible

What happens when:What happens when:Polar covalent molecules are the Polar covalent molecules are the

solute and water is the solvent? solute and water is the solvent? Dissociation does NOT occur. Water is Dissociation does NOT occur. Water is polar and its “oppositely charged polar and its “oppositely charged poles” will be attracted to other polar poles” will be attracted to other polar molecules' “oppositely charged molecules' “oppositely charged poles.” When a solution is made poles.” When a solution is made between two polar molecules it is between two polar molecules it is called molecular solvation.called molecular solvation.

MiscibleMiscible

What happens when:What happens when:Nonpolar covalent molecules are the Nonpolar covalent molecules are the

soluteand water is the solvent? soluteand water is the solvent? A A solution will NOT occur. Water and any solution will NOT occur. Water and any nonpolar molecule will not mix! Think of nonpolar molecule will not mix! Think of putting water and oil together. Water is putting water and oil together. Water is polar and oil is nonpolar. The polar water polar and oil is nonpolar. The polar water is not attracted to the oil, because the is not attracted to the oil, because the nonpolar oil does not have any oppositely nonpolar oil does not have any oppositely charged poles! charged poles!

ImmiscibleImmiscible

IV. Like Dissolves LikeIV. Like Dissolves Like

We don’t always use water as the We don’t always use water as the solvent! Solutions can be made from solvent! Solutions can be made from various substances – a rule of thumb various substances – a rule of thumb to follow when trying to determine if to follow when trying to determine if two substances will form a solution is two substances will form a solution is “like dissolves like.” “like dissolves like.”

Polar Molecules + Polar MoleculesPolar Molecules + Polar Molecules

Nonpolar Molec. + Nonpolar Molec.Nonpolar Molec. + Nonpolar Molec.

Ionic and IonicIonic and Ionic

Ionic + Polar MoleculesIonic + Polar Molecules

Polar Molecules + Nonpolar Molecules Polar Molecules + Nonpolar Molecules

Ionic + Nonpolar MoleculesIonic + Nonpolar Molecules

Water and Alcohol : Miscible

Oil and Hexane : Miscible

Chemical Reaction

Salt and Water : Miscible

Water and Oil : ImmiscibleSalt and Oil: Immiscible

V. Solubility - Now that we can V. Solubility - Now that we can figure out what we can mix to figure out what we can mix to make a solution, how do we make a solution, how do we know how much solvent and how know how much solvent and how much solute to use?much solute to use?

Solubility: Solubility: the maximum amount of a the maximum amount of a substance that will dissolve in a substance that will dissolve in a solvent (at a specific temperature)solvent (at a specific temperature)

According to solubility, solutions can be According to solubility, solutions can be either:either:

unsaturated – unsaturated – a solution that is able to a solution that is able to dissolve more solute (not enough)dissolve more solute (not enough)

saturated –saturated – a solution that cannot a solution that cannot dissolve any more solute (just enough)dissolve any more solute (just enough)

supersaturated –supersaturated – a solution that a solution that contains more solute than can be contains more solute than can be dissolved (too much!!)dissolved (too much!!)

The solubility of substances varies The solubility of substances varies widely. For example 0.189 grams of widely. For example 0.189 grams of Ca(OH)2 dissolves in 100 grams of Ca(OH)2 dissolves in 100 grams of water at 0water at 0C. 122 grams of AgNO3 C. 122 grams of AgNO3 dissolves in 100 grams of water at dissolves in 100 grams of water at 00C. (page 404 in your book)C. (page 404 in your book)

VI. Factors Effecting Rate VI. Factors Effecting Rate and Solubility and Solubility

A.A. Factors Effecting Rate:Factors Effecting Rate:

1. Agitation – stirring or mixing the solution 1. Agitation – stirring or mixing the solution will increase the rate or how fast the will increase the rate or how fast the solute dissolves, but it will not change how solute dissolves, but it will not change how much solute can be dissolved. If you add much solute can be dissolved. If you add 35.9 grams of salt to water (at 2035.9 grams of salt to water (at 20C) it will C) it will all eventually dissolve, but if you stir the all eventually dissolve, but if you stir the solution it will dissolve much quicker. (As solution it will dissolve much quicker. (As you stir the particles are constantly being you stir the particles are constantly being moved, allowing for interactions between moved, allowing for interactions between solute and solvent to occur more quickly.)solute and solvent to occur more quickly.)

2.2. Surface Area – increasing the surface Surface Area – increasing the surface area of the solute will increase the area of the solute will increase the rate or how fast the solute dissolves, rate or how fast the solute dissolves, but it will not change how much but it will not change how much solute can be dissolved.solute can be dissolved.

3. Temperature – increasing 3. Temperature – increasing temperature will increase the rate or temperature will increase the rate or how fast the solute dissolves in the how fast the solute dissolves in the solvent. (As temperature increases solvent. (As temperature increases the particles begin to move faster and the particles begin to move faster and faster and collide with more particles faster and collide with more particles quicker, which means the solute and quicker, which means the solute and solvent particles have an increased solvent particles have an increased chance of coming into contact with chance of coming into contact with each other.)each other.)

B. Factors Effecting B. Factors Effecting Solubility:Solubility:

1. Increasing Temperature - 1. Increasing Temperature - solubility of a solid solute in a liquid solubility of a solid solute in a liquid solvent generally increases with an solvent generally increases with an increase in temperature. At 20increase in temperature. At 20C C 35.9 grams of salt will dissolve in 100 35.9 grams of salt will dissolve in 100 grams of water, but at 100 grams of water, but at 100 C 39.2 C 39.2 grams of salt will dissolve in 100 grams of salt will dissolve in 100 grams of water!grams of water!

2. Decreasing 2. Decreasing Temperature-increases Temperature-increases the solubility of a gaseous the solubility of a gaseous solute in a liquid solvent. solute in a liquid solvent. What would you rather What would you rather drink, a hot coke or a cold drink, a hot coke or a cold coke?coke?

3. Pressure – The solubility of a gas 3. Pressure – The solubility of a gas increases as the pressure of the increases as the pressure of the gas above the liquid increases. gas above the liquid increases. Carbonated drinks have COCarbonated drinks have CO22 dissolved in them. They are also dissolved in them. They are also bottled under a high pressure of bottled under a high pressure of COCO22, which forces the CO, which forces the CO22 into into solution. When the bottle is solution. When the bottle is opened, the pressure above the opened, the pressure above the solution decreases, and bubbles of solution decreases, and bubbles of COCO22 form in the liquid, then escape. form in the liquid, then escape. Eventually, most of the COEventually, most of the CO22 escapes escapes and the drink becomes “flat.”and the drink becomes “flat.”

Henry’s Law- “At a given Henry’s Law- “At a given temperature, the solubility, temperature, the solubility, S, of a gas in a liquid is S, of a gas in a liquid is directly proportional to the directly proportional to the pressure, P, of the gas pressure, P, of the gas above the liquid.”above the liquid.”

S1S1 = = S2 S2 P1 P2P1 P2

VII. Electrolyte VS VII. Electrolyte VS NonelectrolyteNonelectrolyte

1. Electrolyte1. Electrolyte – – compounds that compounds that conduct an electric current in an conduct an electric current in an aqueous solution OR in the molten aqueous solution OR in the molten state. An electrolyte solution contains state. An electrolyte solution contains charged particles (ions), which can charged particles (ions), which can move. Any salt dissolved in water is an move. Any salt dissolved in water is an electrolyte: NaCl, KI, etc. Some polar electrolyte: NaCl, KI, etc. Some polar molecules also conduct electricity (most molecules also conduct electricity (most acids are electrolytes because H is the acids are electrolytes because H is the only nonmetal that has a + charge).only nonmetal that has a + charge).

Types of ElectrolytesTypes of Electrolytes

1. 1. Strong electrolytes Strong electrolytes – a large portion – a large portion of the solute exists as ions: of the solute exists as ions:

a. aqueous solutions of all ionic a. aqueous solutions of all ionic compoundscompounds

b. strong acids: have at least 2 oxygens b. strong acids: have at least 2 oxygens per hydrogen (Hper hydrogen (H22SOSO44, HNO, HNO33))

c. strong bases – these are hydroxides c. strong bases – these are hydroxides from Group I and II, except Be. (NaOH, from Group I and II, except Be. (NaOH, CsOH, etc)CsOH, etc)

2. 2. Weak electrolytesWeak electrolytes – these are solutions – these are solutions in which only a small portion of the solute in which only a small portion of the solute exists exists as ions as ions

a. weak acids:a. weak acids:-all binary acids – HF, H-all binary acids – HF, H22S, etcS, etc-weak acids that have less than 2 oxygen's -weak acids that have less than 2 oxygen's per hydrogen per hydrogen

b. weak bases – hydroxides of everything b. weak bases – hydroxides of everything else not in Group I or II, including Be(OH)else not in Group I or II, including Be(OH)22

2. Non-Electrolytes2. Non-Electrolytes- compounds that - compounds that do NOT conduct electricity in either do NOT conduct electricity in either aqueous solution or when melted:aqueous solution or when melted:

distilled waterdistilled water gasesgases molecular compounds (2 nonmetals)molecular compounds (2 nonmetals) organic compounds – alcohols, sugars, organic compounds – alcohols, sugars,

etc. anything containing a etc. anything containing a CarbonCarbon

Practice Problems: Tell whether each Practice Problems: Tell whether each of the following aqueous solutions of the following aqueous solutions would be a STRONG, WEAK, or NON would be a STRONG, WEAK, or NON electrolyte.electrolyte.

1.1. NaClNaCl 2. CH2. CH33Br (l)Br (l) 3. HMnO3. HMnO44

4.4. HCHC22HH33OO22 5. LiOH5. LiOH 6. HC6. HC66HH77OO66

7. CO7. CO22 (l) (l) 8. HF8. HF

VIII. Concentration: VIII. Concentration: the the concentration of a solution is a concentration of a solution is a measure of the amount of solute in a measure of the amount of solute in a given amount of solvent or solution. given amount of solvent or solution.

1. Molarity: 1. Molarity: the number of moles of the number of moles of solute in one liter of solution.solute in one liter of solution.M = M = amount of solute (moles)amount of solute (moles)

volume of solution (liters)volume of solution (liters)

OROR

M = M = grams of solute/molar mass of grams of solute/molar mass of solute solute

L of solutionL of solution

A bottle labeled 6M HCl is pronounced A bottle labeled 6M HCl is pronounced 6 molar HCl and it was prepared by 6 molar HCl and it was prepared by mixing 6 moles of HCl with enough mixing 6 moles of HCl with enough water to make 1 liter of solution. water to make 1 liter of solution. Molarity is always moles/liter. Ex: 6M Molarity is always moles/liter. Ex: 6M is 6 moles/ literis 6 moles/ liter

6M HCl

Molarity is probably the Molarity is probably the MOST MOST IMPORTANTIMPORTANT unit of concentration unit of concentration that we work with in chemistry. that we work with in chemistry. Knowing the correct technique for Knowing the correct technique for preparing molar solutions is extremely preparing molar solutions is extremely important. A volumetric flask MUST be important. A volumetric flask MUST be used. Here is the technique: (Know it)used. Here is the technique: (Know it)

Determine the correct mass of solute Determine the correct mass of solute needed, add to a volumetric flask and needed, add to a volumetric flask and fill flask with distilled water until it fill flask with distilled water until it reaches the line.reaches the line.

Example Molarity Problems:Example Molarity Problems:1. Calculate the molarity, M, of a solution prepared 1. Calculate the molarity, M, of a solution prepared

by dissolving 11.85 g of potassium permanganate by dissolving 11.85 g of potassium permanganate in enough water to make 750. mL of solution.in enough water to make 750. mL of solution.

2. Calculate the mass of NaCl needed to prepare 2. Calculate the mass of NaCl needed to prepare 175 ml of 0.500 M saline solution.175 ml of 0.500 M saline solution.

3. Calculate the volume (in mL) needed to prepare 3. Calculate the volume (in mL) needed to prepare a 2.48 M sodium hydroxide solution containing a 2.48 M sodium hydroxide solution containing 31.52 g of the dissolved solid.31.52 g of the dissolved solid.

4. How many grams of calcium chloride must be 4. How many grams of calcium chloride must be dissolved in water to make 350. mL of a l.75 M dissolved in water to make 350. mL of a l.75 M solution?solution?

5. What is the molar mass of 55.0 grams of a solute 5. What is the molar mass of 55.0 grams of a solute that has been dissolved in enough solvent to that has been dissolved in enough solvent to make 500. mL of a l.5 M solution?make 500. mL of a l.5 M solution?

2. Molarity of Ions in solution2. Molarity of Ions in solution – – Most ionic solids when dissolved in Most ionic solids when dissolved in water, ionizewater, ionize

Ex: CaClEx: CaCl22 (aq) (aq) Ca Ca2+2+ + 2 Cl + 2 Cl--

NaNa33POPO44 3 Na 3 Na++ + PO + PO443-3-

Mg(OH)Mg(OH)22 Mg Mg2+2+ + 2 OH + 2 OH--

Example Molarity of Ions Problems: Example Molarity of Ions Problems: Calculate the molarity of the ions in Calculate the molarity of the ions in the following solutions:the following solutions:

1. 0.25 M calcium phosphide1. 0.25 M calcium phosphide2. 2.0 M Chromium (III) chloride2. 2.0 M Chromium (III) chloride3. 0.25 M barium hydroxide3. 0.25 M barium hydroxide4. 0.55 M aluminum nitride4. 0.55 M aluminum nitride

3. Dilutions3. Dilutions – you need to make 800.0 mL of a – you need to make 800.0 mL of a 0.25 M solution of HCl. The only available HCl is 0.25 M solution of HCl. The only available HCl is concentrated (12 M). How would you do this? concentrated (12 M). How would you do this? Being able to prepare dilutions is a very common Being able to prepare dilutions is a very common application of chemistry. Our department buys application of chemistry. Our department buys concentrated acids, but normally uses more dilute concentrated acids, but normally uses more dilute solutions of these acids in our labs. Therefore, it solutions of these acids in our labs. Therefore, it is important to know how to correctly dilute. is important to know how to correctly dilute.

MM11VV11 = M = M22VV22

where Mwhere M11 = concentrated solution = concentrated solution VV11 = volume of concentrated solution = volume of concentrated solutionMM22 = dilute solution = dilute solution VV22 = volume of dilute solution = volume of dilute solution

In the above problem, In the above problem, MM11 = 12.0 M M = 12.0 M M22 = 0.25 M = 0.25 M VVll = ? V = ? V22 = 800.0 ml therefore = 800.0 ml therefore (12.0 M) (V(12.0 M) (V11) = (0.25 M)(800 mL) ) = (0.25 M)(800 mL) and Vand V11 = 16.67 ml. = 16.67 ml.This 16.67 mL is the This 16.67 mL is the amount of concentrated amount of concentrated

acid we will take outacid we will take out, but we still have to , but we still have to make 800 mL of 0.25 M. The other 783.3 make 800 mL of 0.25 M. The other 783.3 mL of solution must be water (800 – mL of solution must be water (800 – 16.67). Therefore we would place 16.67 16.67). Therefore we would place 16.67 mL of HCl into 783.3 mL of water and we mL of HCl into 783.3 mL of water and we would have our diluted solution would have our diluted solution

Practice Dilutions Problems:Practice Dilutions Problems:l. How would you prepare 485 mL of 0.39 M l. How would you prepare 485 mL of 0.39 M

solution of NaCl when a l.0 M solution of NaCl solution of NaCl when a l.0 M solution of NaCl is all you can find?is all you can find?

2. If 300.0 mL of a 2.5 M solution of nitric acid 2. If 300.0 mL of a 2.5 M solution of nitric acid is added to 500.0 mL of water, what is the is added to 500.0 mL of water, what is the molarity of the dilute solution?molarity of the dilute solution?

3. Prepare 500. mL of a dilute solution (0.50 M ) 3. Prepare 500. mL of a dilute solution (0.50 M ) of nitric acid from the l5.0 M stock solution. of nitric acid from the l5.0 M stock solution. You will need 600. mL of the dilute solution.You will need 600. mL of the dilute solution.

4. How would you prepare 500. mL of a 0.250 M 4. How would you prepare 500. mL of a 0.250 M solution of NaCl from a 3.00 M stock solution?solution of NaCl from a 3.00 M stock solution?

5. Tell how you would prepare enough 0.75 M 5. Tell how you would prepare enough 0.75 M NaCl solution so that 78 students working in NaCl solution so that 78 students working in groups of 2 will have 12 mL of solution for a groups of 2 will have 12 mL of solution for a lab. The stock solution is 3.5 M NaCl.lab. The stock solution is 3.5 M NaCl.

4. Percent Solutions (2 types):4. Percent Solutions (2 types):1. Percent mass1. Percent mass or or % (m/m)% (m/m)– used – used

when a solid solute is dissolved in when a solid solute is dissolved in liquid, usually water.liquid, usually water.

% (m/m) = % (m/m) = grams of solutegrams of solute grams of solution grams of solution

Example 1: Prepare a 10.00 % NaCl Example 1: Prepare a 10.00 % NaCl solution using 50.0 g water and solid solution using 50.0 g water and solid salt:salt:

0.l0 = x / 50 + x and x = 5.67 g 0.l0 = x / 50 + x and x = 5.67 g NaCl in 50 grams of waterNaCl in 50 grams of water

Example 2: How many grams of water Example 2: How many grams of water must be added to 25.0 g salt in order to must be added to 25.0 g salt in order to have a 2.00 % (by mass) salt solution?have a 2.00 % (by mass) salt solution?

Example 3. Prepare 600.0 g of a 3.00 % Example 3. Prepare 600.0 g of a 3.00 % saline solution (NaCl solution).saline solution (NaCl solution).

2. Percent volume2. Percent volume or or %(v/v) %(v/v) – used – used when a liquid solute is mixed with a when a liquid solute is mixed with a liquid solvent. The units are mL or L, liquid solvent. The units are mL or L, but are worked the same.but are worked the same.

% (v/v) = % (v/v) = mL solutemL solute mL solutionmL solution

Example 1: Prepare a 20.00 % alcohol Example 1: Prepare a 20.00 % alcohol solution using 400.0 mL of water:solution using 400.0 mL of water:

0.20 = x / 400 + x and 0.20 = x / 400 + x and x = 100 mL alcohol + 400 mL waterx = 100 mL alcohol + 400 mL water

Example 2: What is the percent (v/v) Example 2: What is the percent (v/v) of ethanol in the final solution when of ethanol in the final solution when 90.0 mL of it are diluted to a volume 90.0 mL of it are diluted to a volume of 300. mL with water?of 300. mL with water?

5. Molality: 5. Molality: is the concentration of a solution expressed in is the concentration of a solution expressed in moles of solute per kilogram of solvent. Molality is moles of solute per kilogram of solvent. Molality is represented by a lower case m.represented by a lower case m.

molality = molality = amount of solute (moles)amount of solute (moles) mass of solvent (kg)mass of solvent (kg)

OROR

molality = molality = grams of solute/molar massgrams of solute/molar mass kg of solvent kg of solvent

5m NaOH is pronounced 5 molal NaOH solution.5m NaOH is pronounced 5 molal NaOH solution.  

5m = 5m = 5 moles of NaOH5 moles of NaOH 1 kg of water 1 kg of water

5m or 5moles/kg5m or 5moles/kg

Example Molality Problems:Example Molality Problems:1. Calculate the molality of a solution prepared by 1. Calculate the molality of a solution prepared by

dissolving 5.0l g sodium sulfate in 700.0 g water.dissolving 5.0l g sodium sulfate in 700.0 g water.2. Prepare a solution that is 2.50 molal barium 2. Prepare a solution that is 2.50 molal barium

nitrate in 1500. grams of water.nitrate in 1500. grams of water.3. A solution is prepared by dissolving 3.00 g of 3. A solution is prepared by dissolving 3.00 g of

potassium chromate in 58.5 g of water. Calculate potassium chromate in 58.5 g of water. Calculate the molality of the solution.the molality of the solution.

4. How many kg of water must be added to 8.3 g of 4. How many kg of water must be added to 8.3 g of oxalic acid, H2C204, to prepare a 0.050 m oxalic acid, H2C204, to prepare a 0.050 m solution?solution?

IX. Problems involving IX. Problems involving percent solution and densitypercent solution and density 1.1. density – ratio of mass to volume density – ratio of mass to volume

D = m / vD = m / vExample: Calculate the mass of sodium Example: Calculate the mass of sodium

hydroxide in 300.0 ml of solution that hydroxide in 300.0 ml of solution that is 8.00 % NaOH. The density of the is 8.00 % NaOH. The density of the solution is 1.09 g/ml.solution is 1.09 g/ml.

(1) find the number of grams of solution (1) find the number of grams of solution by using the density of the solutionby using the density of the solution

300 mL 1.09 g300 mL 1.09 g = 327 g solution = 327 g solution mLmL(2) find the grams of NaOH by using (2) find the grams of NaOH by using

the percent solution: the percent solution: 0.08 = x / 327 g x = 26.2 g0.08 = x / 327 g x = 26.2 g