mixtures and solutions -...

تستخدم أوراق العمل للمساعدة على أداء األنشطة داخل الصف،

وال تغين عن الكتاب املدرسي

Mixtures and Solutions

Chemistry worksheets, Mixtures and Solutions

pg. 1 http://chemya.weebly.com

Mixtures

Homogeneous mixtures

Mixture: Mixing two or more substances in which each substance retains its chemical

properties .

Homogenous mixture: A mixture whose ingredients cannot be distinguished .

Solute: The substance that dissolves .

Solvent: The medium that dissolves the solute .

Types of solutions

The type of solution depends on the physical state of solute and solvent,

Solute SolventExample Solution type

Oxygen (gas) Nitrogen (gas) Air Gas

Carbon Dioxide (Gas) Water (Liquid) Carbonated water Liquid

Oxygen (Gas) Water (Liquid) Sea water Liquid

Ethylene Glycol (Liquid) Water (Liquid) Antifreeze Liquid

Ethanolic Acid (Liquid) Water (Liquid) Vinegar Liquid

Sodium chloride (solid) Water (liquid) Sea water Liquid

Mercury (liquid) silver (solid) Dental amalgam Solid

Carbon (solid) Iron (solid) Steel Solid

When a substance is dissolved in another, there are two possibilities,

✓ the substance dissolves, in this case the dissolved substance is called soluble substance.

✓ the substance does not dissolve, in this case it is called insoluble substance .

In case of two liquids, the names changed to a miscible and immiscible.

1) Classify the following substances as soluble or insoluble .

Sugar in water __________________________ ______________ ____

Sand in water __________________________ ______________ ____

Oil with water __________________________ ______________ ____

Vinegar with water __________________________ ______________ ____



Heterogeneous mixtures

Heterogeneous mixture: A mixture in which the ingredients are not fully mixed together.

Which can distinguish the components .

Suspension

Suspension: A mixture containing particles that settle out if left undisturbed.

The suspended particle diameters are more than 1000nm, the suspension components are

separated using a filter paper. Such as mud .

Thixotropic suspensions

Suspensions will separate into two distinct layers if left undisturbed for a while—a solid like

substance on the bottom and water on the top. However, when stirred, the solid like substance

quickly begins flowing like a liquid.

On the other hand, there are another type of suspensions which form solid substance in

response to the movement, It can be used in constructions in earthquake areas.

Colloids

Colloid: A heterogeneous mixture made up of medium-volume particles, larger than atoms.

Colloidal particle diameters range from 1nm to 1000nm and are and do not settle out, such as

milk, the components of milk can’t be separated by settling or filtration.

The mixture contains more than one substance, but the quantities of substances are different,

the substance in abundance is called the dispersion medium, while the particles of other

substances (dispersed particles)



Types of colloids

Dispersion medium

Dispersed particles

Example Classification

Solid Solid Colored gemstone Steel in solid

Liquid Solid blood / Gelatin Solid in liquid

solid liquid butter, cheese Solid emulsion

liquid liquid milk, mayonnaise Emulsion

solid gas marshmallow, soaps that float Solid foam

liquid gas whipped cream, beaten egg white Foam

gas solid aerosol smoke, dust in air Solid

What prevent the dispersed particles in a colloid from settling out?

✓ Electrostatic layers around particles

There are charged or polar atomic groups on the surface of

the dispersed particles that attract the positive or negative

charged areas on the dispersion medium. As the electrostatic

layers around the particles repel each other, they remain in

the mixture and do not precipitate .

✓ Brownian motion

Brownian motion: the erratic movement of colloid particles .

Discovered by a Scottish scientist named Robert Brown by the

movement of pollen in water. It produces a collision of the

dispersion medium with the dispersed particles. These collisions prevent particle deposition .

2) List two ways to settle the colloid particles out of the mixture?

1. The stirring of an electrolyte (ionic compound solution) in a colloidal mixture, causes the

molecules to clump together and precipitated in the end .

2. Heating, where heat gives particles enough kinetic energy to overcome electrostatic

forces, causing the precipitation of dispersed particles .



Tyndall effect

The diluted heterogeneous mixture does not appear

cloudy or opaque like concentrated colloids, but it

does not pass light as efficiently as the homogeneous

solutions, we can see this clearly when a beam of

sunlight passes through fog or clouds .

Tyndall effect: Scattering the light due to the

presence of dispersed particles or suspended in the

heterogeneous mixture .

3) What is the difference between a concentrated heterogeneous solution and a diluted

heterogeneous solution when trying to pass light

through them?

The heterogeneous solution appear as opaque

substance, which does not pass the light, while the

diluted mixture appears transparent, passing light but

dispersing it .

4) Why does the diluted mixture scatter the light while concentrated mixture appears

opaque?

Due to the low number of particles dispersed in the diluted mixture .

5) Compare between the homogeneous, collide, and suspension mixture

Suspension colloid Homogeneous mixture

Definition

Particle volume

Tyndall

phenomenon

Components

precipitation

Example



Define the following,

6) Solvent: __________________________________________________________________

7) Solute: ___________________________________________________________________

8) Dispersion medium: ________________________________________________________

9) Dispersion medium: ________________________________________________________

10) Why does it more difficult to drive during fog with high lights than using low lights? _____________________________________________________________________________________________________________________________________

_____________________________________________________________________________________________________________________________________

_____________________________________________________________________________________________________________________________________

11) Explain why the dissolved particles remain dispersed in the colloidal mixture . _____________________________________________________________________________________________________________________________________

_____________________________________________________________________________________________________________________________________

_____________________________________________________________________________________________________________________________________

12) What causes Brownian motion? _____________________________________________________________________________________________________________________________________

_____________________________________________________________________________________________________________________________________

_____________________________________________________________________________________________________________________________________



Solution Concentration

Concentration: A measure of how much solute is dissolved in a specific amount of solvent .

The concentration can be expressed qualitatively using a

diluted term or concentrated,

Dilute solution: A solution that contains a small amount of

solute .

Concentrated solution: A solution containing a greater

amount of solute .

Among the methods of calculating the concentration is the molarity, which is preferred by

chemists, mention the reason .

Molarity (Molar concentration) =solute moles number

solution volume in liters

The equation shows the number of solute moles and the number of particles, so chemists

prefer the use of molar .

Quantitative calculation of concentration

Percent by mass

Is the ratio of the solute mass to the mass of the solution expressed as a percent.

The mass of the solution is equal to the sum of dissolved solid and solvent.

Percent by mass = 𝒎𝒂𝒔𝒔 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒆

𝒎𝒂𝒔𝒔 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 × 100

13) A sample of 3.6 g table salt NaCl added to 100g of water, Calculate the percentage by mass

of NaCl salt .

14) What is the percent by mass of NaHCO3 in a solution containing 20.0 g of NaHCO3

dissolved in 600.0 mL of H2O?

_____________________________________________________________________________________________________________________________________

_____________________________________________________________________________________________________________________________________

_____________________________________________________________________________________________________________________________________

_____________________________________________________________________________________________________________________________________

Mass dissolved = 3.6g

Solvent mass = 100g

Percentage =?

Solution mass = Solvent mass + Solute mass

3.6 g + 100.0 g = 103.6 g

Percent by mass = 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒

𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 × 100

= 3.6 g

103.6 g × 100 = 3.5%



15) You have 1500.0 g of a bleach solution. The percent by mass of the solute sodium

hypochlorite (NaOCl) is 3.62%. How many grams of NaOCl are in the solution?

_____________________________________________________________________________________________________________________________________

_____________________________________________________________________________________________________________________________________

_____________________________________________________________________________________________________________________________________

_____________________________________________________________________________________________________________________________________

16) In previous question, how many grams of solvent are in the solution?

_____________________________________________________________________________________________________________________________________

_____________________________________________________________________________________________________________________________________

_____________________________________________________________________________________________________________________________________

_____________________________________________________________________________________________________________________________________

17) The percent by mass of calcium chloride in a solution is found to be 2.65%. If 50.0 g of

calcium chloride is used, what is the mass of the solution?

_____________________________________________________________________________________________________________________________________

_____________________________________________________________________________________________________________________________________

_____________________________________________________________________________________________________________________________________

_____________________________________________________________________________________________________________________________________

Percent by volume

the ratio of solute to solution volume, usually used to calculate mixed fluid ratios .

Biodiesel is a clean-burning alternative fuel that is produced from renewable resources.

Biodiesel can be used in diesel engines with little or no modifications.

18) What are the advantages of Biodiesel?

Biodiesel is simple to use, biodegradable, nontoxic, and it does not contain sulfur or aromatics.

It does not contain petroleum, but it can be blended with petroleum diesel to create a

biodiesel blend. B20 is 20% by volume biodiesel, 80% by volume petroleum diesel.

percentage by volume = 𝒔𝒐𝒍𝒖𝒕𝒆

𝒗𝒐𝒍𝒖𝒎𝒆 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 100 ×

19) What is the percent by volume of ethanol in a solution that contains 35 mL of ethanol

dissolved in 155 mL of water?

_____________________________________________________________________________________________________________________________________

_____________________________________________________________________________________________________________________________________

_____________________________________________________________________________________________________________________________________

_____________________________________________________________________________________________________________________________________



20) What is the percent by volume of isopropyl alcohol in a solution that contains 24 mL of

isopropyl alcohol in 1.1 L of water?

_____________________________________________________________________________________________________________________________________

_____________________________________________________________________________________________________________________________________

_____________________________________________________________________________________________________________________________________

21) If 18 mL of methanol is used to make an aqueous solution that is 15% methanol by volume,

how many milliliters of solution is produced?

_____________________________________________________________________________________________________________________________________

_____________________________________________________________________________________________________________________________________

_____________________________________________________________________________________________________________________________________

Molarity (Molar concentration) (M)

The number of moles of solute per liter of solution.

(M)= 𝒎𝒐𝒍𝒆𝒔 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒆

𝑺𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝑽𝒐𝒍𝒖𝒎𝒆(𝑳)

When one liter of a solution contains (1mol) one mol of the solute the concentration is 1 mol

1L

, 1M (one molar)

22) Determine What is the molar concentration of a liter solution with 0.5 mol of solute?

Number of solute moles = 0.5 mol

Solution volume = 1.0 L

The molarity =?

(M)= 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒

𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑉𝑜𝑙𝑢𝑚𝑒(𝐿)

0.5mol

1.0L= 0.5M (0.5 Molar)

23) A 100.5-mL intravenous (IV) solution contains 5.10 g of glucose (C6H12O6 ). What is the

molarity of this solution? The molar mass of glucose is 180.16 g/mol.

Mass dissolved = 5.10 g

Solution volume = 100.5 mL

The molarity =?

1) Calculate the number of dissolved moles

5.10g C6H12O6 × 1𝑚𝑜𝑙

180.16𝑔 = 0.0283 mol C6H12O6

2) Convert volume to liter

100.5 mL × 1𝐿

1000 𝑚𝐿 = 0.1005 L

3) Concentration calculations

(M)= 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒

𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑉𝑜𝑙𝑢𝑚𝑒(𝐿)



0.0283 mol0.1005 L

= 0.282 M

24) What is the molarity of an aqueous solution containing 40.0 g of glucose (C6H12O6 ) in 1.5L

of solution?

_____________________________________________________________________________________________________________________________________

_____________________________________________________________________________________________________________________________________

_____________________________________________________________________________________________________________________________________

_____________________________________________________________________________________________________________________________________

25) Calculate the molarity of 1.60L of a solution containing 1.55g of dissolved KBr.

_____________________________________________________________________________________________________________________________________

_____________________________________________________________________________________________________________________________________

_____________________________________________________________________________________________________________________________________

_____________________________________________________________________________________________________________________________________

26) What is the molarity of a bleach solution containing 9.0 g of NaOCl per liter of bleach?

_____________________________________________________________________________________________________________________________________

_____________________________________________________________________________________________________________________________________

_____________________________________________________________________________________________________________________________________

_____________________________________________________________________________________________________________________________________

27) How much calcium hydroxide (Ca(OH)2), in grams, is needed to produce 1.5L of a 0.25M

solution?

_____________________________________________________________________________________________________________________________________

_____________________________________________________________________________________________________________________________________

_____________________________________________________________________________________________________________________________________

_____________________________________________________________________________________________________________________________________



Preparing Molar solutions

To prepare a 1L water solution of 1.50 M concentration of CuSO4.5H2O copper II sulfate, follow

these steps :

1. calculate the mass of copper sulphate necessary to prepare 1.50 mol

1.50 mol CuSO4. 5H2O × 249.7 𝑔

1 𝑚𝑜𝑙 = 375 g CuSO4. 5H2O

2. Put the previous mass 375 g in a small amount of water less than 1L and dissolve it until

dissolving completely .

3. Complete the solution with water until it reaches 1L

In our reactions we do not necessarily use all the previous quantity. A reaction may require

only 100mL of substance, so the required quantity should only be prepared so as not to waste

large quantities of chemicals. To prepare less than 1L of substance we use other calculations .

To prepare 100 mL of 1.50 M CuSO4.5H2O which is the same as the previous concentration but

less quantity we follow the following steps,

1. calculate the mass of copper sulphate necessary to prepare 1.50 mol in 1L 1.50 mol CuSO4.5H2O

1L ×

249.7 𝑔

1 𝑚𝑜𝑙 =

375 g CuSO4.5H2O

1L

2. convert the appropriate volume from 1L to 100 mL 375 g CuSO4.5H2O

1L× 100 mL ×

1𝐿

1000 𝑚𝐿 = 37.5 g CuSO4.5H2O

3. Dissolve the previous mass 37.5 g CuSO4.5H2O in a small amount of water and then fill

it up to 100 mL



28) How many grams of CaCl2 would be dissolved in 1.0 L of a 0.10M solution of CaCl2?

_____________________________________________________________________________________________________________________________________

_____________________________________________________________________________________________________________________________________

_____________________________________________________________________________________________________________________________________

_____________________________________________________________________________________________________________________________________

29) How many grams of CaCl2 should be dissolved in 500.0 mL of water to make a 0.20M

solution of CaCl2?

_____________________________________________________________________________________________________________________________________

_____________________________________________________________________________________________________________________________________

_____________________________________________________________________________________________________________________________________

_____________________________________________________________________________________________________________________________________

30) How much NaOH are in 250 mL of a 3.0M NaOH solution?

_____________________________________________________________________________________________________________________________________

_____________________________________________________________________________________________________________________________________

_____________________________________________________________________________________________________________________________________

_____________________________________________________________________________________________________________________________________

31) What volume of ethanol (C2H3OH) is in 100.0 mL of 0.15M solution? The density of ethanol

is 0.7893 g/mL.

_____________________________________________________________________________________________________________________________________

_____________________________________________________________________________________________________________________________________

_____________________________________________________________________________________________________________________________________

_____________________________________________________________________________________________________________________________________



Diluting molar solutions

We usually use chemicals in specific quantities that are known. These quantities may vary

between mass, volume, number of moles or concentration. These quantities are called

standard quantities, which are constant in all laboratories and are used for measuring other

quantities. One of the most commonly known substances is HCl hydrochloric acid standard

solution, its concentration is 12M

It is not necessarily necessary to use the solution in the laboratory with the same

concentration. We may need a solution containing less concentration, so we will have to dilute

the solution. To do this we add a quantity of solvent, and the important question is what is the

appropriate amount of concentrated solution that we need to use to add the solvent?

In diluting process, we add an amount of solvent to specific amount of the solution, so what

changes is the volume of the solution, on the other hand the number of moles of solute still

the same. We calculate the number of moles of solute in the solution from the equation

(M)= 𝒎𝒐𝒍𝒆𝒔 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒆

𝑺𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝑽𝒐𝒍𝒖𝒎𝒆(𝑳)

𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 = Molarity × 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑉𝑜𝑙𝑢𝑚𝑒(𝐿)

Case 1 (before diluting) n1 = M1 × V1

Case 2 (after diluting) n2 = M2 × V2

Because the number of moles of solute doesn’t change, so n1 = n2 so,

Dilution Equation

M1V1=M2V2 M1 Molarity before diluting

V1 Volume before diluting

M2 Molarity after diluting

V2 Volume after diluting



32) What volume, in milliliters, of 2.00M calcium chloride (CaCl2) stock solution would you use

to make 0.50 L of 0.300M calcium chloride solution?

M1 = 2.00 M CaCl2

V1 = ?

M2 = 0.300 M CaCl2

V2 = 0.50 L

M1V1=M2V2

V1 = 𝑀2𝑉2

𝑀1

V1 = 0.300 M × 0.50 L

2.00 M = 0.075 L

V1 =0.075 L × 1000 mL

1L = 75 mL

33) What volume of a 3.00M KI stock solution would you use to make 0.300L of a 1.25M KI

solution?

_____________________________________________________________________________________________________________________________________

_____________________________________________________________________________________________________________________________________

_____________________________________________________________________________________________________________________________________

_____________________________________________________________________________________________________________________________________

34) How many milliliters of a 5.0M H2SO4 stock solution would you need to prepare 100.0mL

of 0.25M H2SO4?

_____________________________________________________________________________________________________________________________________

_____________________________________________________________________________________________________________________________________

_____________________________________________________________________________________________________________________________________

_____________________________________________________________________________________________________________________________________

35) If 0.5L of 5M stock solution of HCl is diluted to make 2L of solution, how much HCl, in grams,

was in the solution?

_____________________________________________________________________________________________________________________________________

_____________________________________________________________________________________________________________________________________

_____________________________________________________________________________________________________________________________________

_____________________________________________________________________________________________________________________________________



Molality (molal concentration)

Number of dissolved moles of solute in a specific mass of solvent .

(m)= 𝒎𝒐𝒍𝒆𝒔 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒆

𝒔𝒐𝒍𝒗𝒆𝒏𝒕 𝒎𝒂𝒔𝒔(𝒌𝒈)

36) Why do chemists prefer molality than molarity?

Because molarity changes by changing temperature, as the temperature increase the solution

volume expands so the concentration decreases.

In the lab, a student adds 4.5 g of sodium chloride (NaCl) to 100.0 g of water. Calculate the

molality of the solution.

solute mass = 4.5g NaCl

Solvent mass = 100.0g

Molality (m) =?

4.5g NaCl × 1𝑚𝑜𝑙 𝑁𝑎𝐶𝑙

58.44 𝑔 𝑁𝑎𝐶𝑙 = 0.077 mol NaCl

(m)= 𝒎𝒐𝒍𝒆𝒔 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒆

𝒔𝒐𝒍𝒗𝒆𝒏𝒕 𝒎𝒂𝒔𝒔(𝒌𝒈)

100.0 g H2O × 1 𝑘𝑔

1000 𝑔 = 0.1000 kg H2O

m = 0.077 𝑚𝑜𝑙 𝑁𝑎𝐶𝑙

0.1000 𝑔 𝐻2𝑂 = 0.77 mol/kg

37) What is the molality of a solution containing 10.0 g of Na2SO4 dissolved in 1000.0 g of

water?

_____________________________________________________________________________________________________________________________________

_____________________________________________________________________________________________________________________________________

_____________________________________________________________________________________________________________________________________

_____________________________________________________________________________________________________________________________________

38) How much (Ba(OH)2), in grams, is needed to make a 1.00m aqueous solution?

_____________________________________________________________________________________________________________________________________

_____________________________________________________________________________________________________________________________________

_____________________________________________________________________________________________________________________________________

_____________________________________________________________________________________________________________________________________



Mole fraction

the ratio of the number of moles of solute in solution to the total number of moles of solute

and solvent.

39) Example: 100g of hydrochloric acid contains 36g HCl and 64g H2O calculate the mole

fraction of this solution .

Solute mass = 36g HCl

Solvent mass = 64g H2O n HCl = 36g HCl ×

1𝑚𝑜𝑙 HCl

36.5 𝑔 HCl = 0.99 mol HCl

nH2O = 64g H2O × 1 𝑚𝑜𝑙

18.02 𝑔 = 3.6 mol H2O

𝑋𝐻𝐶𝑙 =𝑛𝐻𝐶𝑙

nHCl + nH2O =

0.99 mol HCl

0.99 mol HCl+ 3.6 mol 𝐻2𝑂 = 0.22

𝑋H2O =nH2O

nHCl + nH2O =

3.6 mol 𝐻2𝑂

0.99 mol HCl+ 3.6 mol 𝐻2𝑂 = 0.78

To make sure that the answer is correct, the sum of the molar fraction values should equal

to (1)

0.78 + 0.22 = 1

40) What is the mole fraction of NaOH in an aqueous solution that contains 22.8% NaOH by

mass?

_____________________________________________________________________________________________________________________________________

_____________________________________________________________________________________________________________________________________

_____________________________________________________________________________________________________________________________________

_____________________________________________________________________________________________________________________________________

41) If the mole fraction of sulfuric acid (H2SO4) in an aqueous solution is 0.325, how much

water, in grams, is in 100 mL of the solution?

_____________________________________________________________________________________________________________________________________

_____________________________________________________________________________________________________________________________________

_____________________________________________________________________________________________________________________________________

_____________________________________________________________________________________________________________________________________

Mole fraction

𝑋𝐴 =𝑛𝐴

𝑛𝐴+ 𝑛𝐵 𝑋𝐵 =

𝑛𝐵

𝑛𝐴+ 𝑛𝐵

XA Mole fraction XB Mole fraction nA Solute moles

nB Solvent moles



Factors Affecting Solubility

forces between particles

Attractive forces exist among the particles of all substances.

➢ Attractive forces exist between the pure solute particles.

➢ Attractive forces exist between the pure solvent particles.

➢ Attractive forces exist between the solute and solvent particles.

When mixing solute substances and solvent, there are two options,

✓ The first option is that the attraction forces between particles of solvent and solute

particles are stronger than the attraction forces between solute particles with each

other. Here the solute particles are dissolved and dispersed in the solution .

✓ The second option is that the attraction forces between the solute particles are greater

than the attraction forces between the solvent and solute particles. Here the solute

particles are not separated and remain coherent, in which case the substance does not

dissolve .

Solvation: Surrounding solute particles by solvent particles .

Hydration: Solvation in water .

“Like dissolves like” is the general rule used to determine whether the substance will dissolve

or not.

To determine whether a solvent and solute are alike, you must examine

1. The bonding and polarity of the particles.

2. The intermolecular forces among particles.



Aqueous solutions of ionic compounds

Water molecules consist of dipoles

on oxygen atoms and hydrogen

atoms due to the electronegativity

difference between the atoms. A

partial negative charge is formed on

the oxygen atom while a partial

positive charge is formed on the

hydrogen atom .

When dissolving salt in water, water molecules accumulate in the direction of oxygen

atoms with a partial negative charge around the positively charged sodium ions and

attract them and vice versa with negative chloride ions,

The attraction forces between ions and water molecules are stronger than the attraction

forces between ions. The ions are separated from the crystal and dissolved in water .

If the attraction between the ions of the substance is greater than the attraction

between water molecules and the ions, these ions do not separate as in the case of

plaster made out of gypsum used in the manufacture of the medical trowel .



Aqueous solutions of molecular compounds

42) How does non-polar sugar dissolve in polar water?

The sugar molecule contains many

OH groups and these groups are

polar, because of the difference in

electronegativity between hydrogen

and oxygen (we previously learned

that the molecule may be non-polar

even if some of its bonds are polar)

Hydrogen bonds are formed between water molecules and OH groups in sugar, the

attraction between water and sugar molecules is strong at these points and is greater

than the attraction between sugar molecules, so the sugar molecules break down and

dissolve in water.

Non-polar substances dissolve better in non-polar solvents.

Heat of solution

We always emphasize basic points,

We often require energy to break bonds or to break the attraction between different

molecules.

Energy is often released when bonds are formed or when two different molecules

interact.

Depending on the difference in energy the type of solution (endothermic or exothermic)

If the energy released by the attraction between the solvent molecules and the

dissolved molecules is greater than the energy absorbed by breaking the attraction

between the solvent molecules and also between the solute molecules, so the solution

will be exothermic.

The cup temperature increases when calcium chloride is dissolved in water, which

means that it is exothermic solution.

If the energy released as a result of the attraction between the solvent molecules and

the dissolved molecules is less than the absorbed by breaking the attraction between

the solvent molecules and also between the solute molecules, so the solution will be

endothermic.

The solution temperature decreases when the ammonium nitrate is dissolved in water,

which means that the solution is endothermic.

The heat of solution: The overall energy change that occurs during the solution formation

process.



Factors Affecting solvation

Agitation

The stirring of the solution moves a larger number of solvent particles onto the solute surface,

allowing more collisions or collisions, which accelerate the dissolving process .

Surface area

There is a difference between using a sugar cube and a teaspoon of powdered sugar to dissolve

it in water, where the crushed or ground sugar dissolved faster because of increasing the

surface area exposed to the liquid .

Temperature

The rete of solvation is affected by temperature,

For example, the solvation of solids increases as the temperature of the solution

increases, because increasing the temperature leads to increased particle movement,

causing more collisions between solvent and soluble substances, increasing the chances

of contact between them and increasing their solvation .

In contrast, the solvation of the gaseous substances decreases when the temperature is

raised, because that they acquire a certain amount of energy and it is easy to convert to

gaseous state and to escape from the attraction of the liquid particles .



Solubility

Over time, the amount of dissolved substance increases in the

solution, as the particles of the solvent collide with solute

particles, the dissolved particles collide with each other and

sometimes return to the crystal, causing crystallization .

As the solvation increases, the crystallization rate increases until

reaching equilibrium between solvation and crystallization.

Unsaturated solutions

Is a solution that contains a small amount of solute, so there is no enough collisions between

the solute particles to crystallize again, there are other opportunities in the solution of the

solute particles to be present in the solution, which can dissolve additional quantities .

Saturated solutions

When we add large amounts of solute, the solution eventually reaches the state of equilibrium,

where the amount of dissolved solute in the solution is constant. even we add more solid

solute no more dissolved, at constant pressure and temperature .

Supersaturated solutions

When heating a saturated solution, the kinetic energy of the particles increases, giving solvent

particles another opportunity to collide with solute particles and this causes dissolving of

another amount of solute, here the solution becomes supersaturated with amount of

dissolved particles greater than can be dissolved at normal temperature and pressure .

A supersaturated solution: contains more dissolved solute than a saturated solution at the

same temperature.

Dissolved Calcium chloride (CaCl2) increases

from 64g CaCl2 per 100g of water at a

temperature of 10oC to 100 g CaCl2 when the

temperature is raised to 27oC

Solubility of Some substances such as Cerium

sulphate Ce2(SO4)3 reduces with a temperature

increase .



43) How to prepare a supersaturated solution?

1. Quantities of solute are added until the saturation is reached .

2. Add additional amount of solute and then heat the solution until the additional amount

of solute dissolves .

3. Cool the solution slowly.

44) How can the increase of solute be precipitated in supersaturated solution?

➢ Adding a small amount of solute to the solution, it precipitates very quickly around the

small added part called the seed crestal .

Silver iodide is used as a seed crestal for water vapor in the air, allowing rainfall in a process

called cloud seeding .

➢ Scratching the inner part of the container .

➢ When the supersaturated solution is undergoes a sudden shock such as stirring.



45) There are forces of attraction between different molecules, when dissolving substance in

another one there are two possibilities, mention what is produced if,

The attraction forces between the solute

and solvent particles are smaller than

the forces of attraction between solute

particles

The attraction forces between the solute

and solvent particles are greater than the

forces of attraction between solute

particles

46) Using periodic table Select which atoms of the two elements has more electronegativity

than the other, oxygen O and hydrogen H? And mention partial charge of each.

• The largest ___________________________ partial charge ___________________________

• The smallest __________________________ partial charge ___________________________

47) When dissolving salt in water,

• Which atoms in water molecule will gather around chloride ion? ___________________________

• Which atoms in water molecule will gather around the sodium ion? ___________________________

48) With your information on dissolving salt in water, complete the following sentence with

one of the following (smaller - larger)

The attraction forces between chloride and sodium ions on one side and water on the other

side ___________________________ the forces of attraction between sodium ions and chloride ions.

49) Why does the gypsum used in making the trowel not dissolved in water? _____________________________________________________________________________________________________________________

_____________________________________________________________________________________________________________________

50) What is the type of bond that causes sugar to dissolve in water? explain. _____________________________________________________________________________________________________________________

_____________________________________________________________________________________________________________________

51) Compare the endothermic solution with the exothermic solution.

exothermic solution endothermic solution

___________________________________ ___________________________________

Energy released as a result of the

attraction between solvent and

solute molecules

___________________________________ ___________________________________ Absorbed energy due to broken

bonds between solute particles

___________________________________ ___________________________________ Example



Solubility of gases

The higher the temperature, the higher the particle's kinetic energy, giving the gas molecules

more opportunities to escape from the liquid, thus reduces the solubility of gases as the

temperature increases .

Henry's Law

Gas solubility in liquid directly proportional to pressure at stable temperature .

Bottles of carbonated beverage are filled with carbon dioxide gas at high pressure. When the

package is opened, the pressure on the surface is reduced and the dissolved gas molecules are

released from the liquid .

Henry’s law states that at a given temperature, the solubility (S) of a gas in a liquid is directly

proportional to the pressure (P) of the gas above the liquid.

52) If 0.85 g of a gas at 4.0 atm of pressure dissolves in 1.0 L of water at 25°C, how much will

dissolve in 1.0 L of water at 1.0 atm of pressure and the same temperature?

S1 = 0.85 g/L

P1 = 4.0 atm

S2 = ?

P2 = 1.0 atm

S1

P1

=S2

P2

𝑆2 = S1 × P2

P1

𝑆2 =0.85 g

1.0 𝐿 ×

1.0 atm

4.0 atm = 0.21 g/L

Henry’s law

𝐒𝟏

𝐏𝟏=

𝐒𝟐

𝐏𝟐 S solubility

P pressure

Solubility units (g/L)



53) If 0.55 g of a gas dissolves in 1.0 L of water at 20.0 kPa of pressure, how much will dissolve

at 110.0 kPa of pressure?

_____________________________________________________________________________________________________________________________________

_____________________________________________________________________________________________________________________________________

_____________________________________________________________________________________________________________________________________

_____________________________________________________________________________________________________________________________________

54) A gas has a solubility of 0.66 g/L at 10.0 atm of pressure. What is the pressure on a 1.0-L

sample that contains 1.5 g of gas?

_____________________________________________________________________________________________________________________________________

_____________________________________________________________________________________________________________________________________

_____________________________________________________________________________________________________________________________________

_____________________________________________________________________________________________________________________________________

55) The solubility of a gas at 7 atm of pressure is 0.52 g/L. How many grams of the gas would

be dissolved per 1 L if the pressure was raised to 10 atm?

_____________________________________________________________________________________________________________________________________

_____________________________________________________________________________________________________________________________________

_____________________________________________________________________________________________________________________________________

_____________________________________________________________________________________________________________________________________

mixtures and solutions -...

Documents