Section 18.1: Properties of Solutions

Section 18.2: Concentrations of Solutions

Section 18.3: Colligative Properties of Solutions

Section 18.4: Calculations Involving Colligative Properties

Section 18.1

Recollect that a solution is made up from a solute dissolved in a

solvent.

Solutions are homogeneous mixtures that may be solid,

liquid, or gaseous.

Solutes dissolved in solvents experience interactions at the surface of each, where the solute contacts the solvent.

The rate at which sugar will dissolve in tea is impacted by many factors.

The speed at which a solute (sugar) dissolves is dependent upon the rate at which new solvent

molecules (tea) come into contact with solute molecules.

The rate of this contact can be increased with stirring, or heating the solution mixture.

The rate of this contact can be increased with:

1. Stirring or agitation

2. Increased temperature (kinetic energy)

3. Particle size – smaller ones dissolve more quickly

All solutions have a limit as to the amount of solute that can be dissolved in the solvent.

If you add 36.0 g of NaCl to 100 g of water at 25 ̊C, all 36.0 g dissolves. Add one more gram and stir, no matter

how vigorously, 0.8 g will fall out of the solution.

We would call a solution such as this saturated. No more salt can be dissolved

in it – any additional salt will settle to the bottom.

All solutions have a limit as to the amount of solute that can be dissolved in the solvent.

If you add 36.0 g of NaCl to 100 g of water at 25 ̊C, all 36.0 g dissolves. Add one more gram and stir, no matter

how vigorously, 0.8 g will fall out of the solution.

Note from the figure that as ions are taken up in the

solution, others crystallize, in a

dynamic equilibrium.

A saturated solution contains the maximum amount of solute for a given amount of solvent at a constant

temperature. 36.2 g NaCl in 100 g of water at 25 ̊C produces a

saturated solution.

A solution containing less

than this amount is said to be unsaturated.

Changing the

temperature may affect

the solubility of a

substance.

Notice that increasing the temperature

greatly increases the solubility of KNO3, but

decreases the solubility of Yb2(SO4)3.

Generally, increasing

the temperature

of a solid solute,

increases its solubility.

Try this for yourself, make a saturated sugar solution for

hummingbird food:

Dissolve ¼ cup (56 g) of table sugar in one cup of cold

water (240 mL). What do you notice? The sugar does not

completely dissolve.

Now, place the sugar/ water

solution in the microwave and

heat to almost boiling.

Remove carefully and stir.

What do you notice?

Gases have different

solubilities in water at different

temperatures, as well.

Generally, as the

temperature increases, the solubilities of

gases decreases.

Nitrogen, oxygen and nitric oxide

become virtually

insoluble as the

temperature of a solution

reaches 100 ̊C.

You can taste the difference in carbon dioxide

solubility whenever you drink an open soda, that

has gotten warm.

Warm soda’s are not only

distasteful because they are

warm, but also because they

have lost carbonation.

Various conditions of pressure can also impact the

solubility of a gas dissolved in a 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.

As the pressure of the gas above the liquid increases,

the solubility of the gas dissolved increases.

𝑆1

𝑃1=

𝑆2

𝑃2

A solution that holds more solute than it should

theoretically at a given temperature is

supersaturated.

Below, note the supersaturated solution, before a

seed crystal is added (left).

The same solution after a seed crystal is added

(center).

The seed crystal causes the excess solute to

crystallize rapidly (right).

Section 18.2

Dilute solutions – contain only a low concentration of solute.

Concentrated solutions – contain high

concentrations of solute.

Chemists have to be able to quantify the concentrations of their solutions.

It is not enough to know that a solution is “dilute,” or “concentrated.”

Molarity (M) is the most important unit of concentration in chemistry.

Defines the number of moles of a solute dissolved per liter of solution.

Also called the molar concentration.

A solution has a volume of 2.0 L and contains 36.0 g of

glucose. If the molar mass of glucose is 180 g/ mol, what

is the molarity of the solution?

𝑀𝑜𝑙𝑎𝑟𝑖𝑡𝑦 𝑀 = 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒

𝑙𝑖𝑡𝑒𝑟𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛

Convert 36.0 g of glucose (C6H12O6) to moles first:

36.0 𝑔 𝑔𝑙𝑢𝑐𝑜𝑠𝑒1 𝑚𝑜𝑙 𝑔𝑙𝑢𝑐𝑜𝑠𝑒

180.2 𝑔 𝑔𝑙𝑢𝑐𝑜𝑠𝑒= 0.200 𝑚𝑜𝑙 𝑔𝑙𝑢𝑐𝑜𝑠𝑒

𝑀 =0.200 𝑚𝑜𝑙 𝑔𝑙𝑢𝑐𝑜𝑠𝑒

2.0 𝐿 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛= 0.10 𝑀

A solution has a volume of 250 mL, and contains

0.70 mol NaCl. What is its molarity?

𝑀 = 0.70 𝑚𝑜𝑙 𝑁𝑎𝐶𝑙

0.250 𝐿 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛= 2.8𝑀

How many moles of ammonium nitrate are in

335 mL of 0.425M NH4NO3?

use 𝑀 = 𝑚𝑜𝑙𝑒𝑠 𝑠𝑜𝑙𝑢𝑡𝑒

𝐿 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛; 𝑟𝑒𝑎𝑟𝑟𝑎𝑛𝑔𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑡𝑜:

𝑚𝑜𝑙𝑒𝑠 𝑠𝑜𝑙𝑢𝑡𝑒 = 𝑀 𝐿 𝑠𝑜𝑙𝑛

𝑚𝑜𝑙𝑒𝑠 𝑠𝑜𝑙𝑢𝑡𝑒 =0.425 𝑚𝑜𝑙 𝑁𝐻4𝑁𝑂2

1 𝐿 𝑠𝑜𝑙𝑛0.335 𝐿 𝑠𝑜𝑙𝑛

= 0.142 𝑚𝑜𝑙

Chemists often need to make dilutions of solutions, from stock solutions on their shelves.

When purchasing hydrochloric acid, it comes in a

stock solution concentration of 12M.

Sulfuric acid comes in a stock solution concentration of 6M.

These concentrations are unworkable in everyday

reactions, chemists often have to dilute these significantly.

Suppose you need 250 mL of 0.20M NaCl, but the only

supply of sodium chloride you have is a solution of 1.0M

NaCl. How do you prepare the required solution? Assume

that you have the appropriate volume-measuring devices

on hand.

You will use a measured amount of the stock 1.0M NaCl

solution, and dilute it with enough water to make a total of

0.250 L.

How much of the stock solution do you need?

Suppose you need 250 mL of 0.20M NaCl, but the only

supply of sodium chloride you have is a solution of 1.0M

NaCl. How do you prepare the required solution? Assume

that you have the appropriate volume-measuring devices

on hand.

Use M1V1 = M2V2

Where M1 is the molarity of the stock solution, V1 is the

volume of stock solution you need, M2 is the molarity of

the wanted solution, and V2 is the volume of the solution

you want.

Suppose you need 250 mL of 0.20M NaCl, but the only

supply of sodium chloride you have is a solution of 1.0M

NaCl. How do you prepare the required solution? Assume

that you have the appropriate volume-measuring devices

on hand.

For this problem, you need to solve for V1:

𝑉1 = 𝑀2𝑉2

𝑀1=

(0.20𝑀)(0.250𝐿)

1.0𝑀= 0.050𝐿 𝑠𝑡𝑜𝑐𝑘 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛

If both the solute and the solvent are liquids, a convenient way to make a solution is to

measure volumes.

The concentration of the solute (the smaller quantity liquid) is then

expressed as a percent of the solution by volume.

Isopropyl (rubbing) alcohol that is 70% by volume, is an example of this.

The relationship between percent by volume and the volumes of solute and solution is:

Percent by volume (%(v/v)) =

𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒

𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑣𝑜𝑙𝑢𝑚𝑒x 100%

If 10 mL of pure acetone is diluted with water to a

total solution volume of 200 mL, what is the

percent by volume of acetone in the solution?

Use % by volume = v/ v x 100%:

%𝑣

𝑣=

10 𝑚𝐿 𝑎𝑐𝑒𝑡𝑜𝑛𝑒

200 𝑚𝐿 𝑠𝑜𝑙𝑛 𝑥 100 =

5% 𝑏𝑦 𝑣𝑜𝑙𝑢𝑚𝑒

A bottle of hydrogen peroxide antiseptic is labeled

3.0% (v/v). How many mL H2O2 are in a 400.0-

mL bottle of this solution?

Rearrange equation % = v/v x 100%

solving for volume of solute:

𝑣𝑜𝑙𝑢𝑚𝑒 𝑠𝑜𝑙𝑢𝑡𝑒 =(3.0%)(400.0 𝑚𝐿 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛)

100%

= 12 𝑚𝐿 𝑠𝑜𝑙𝑢𝑡𝑒

If the solute is a solid dissolved in a liquid solvent, it’s percent composition is represented

as mass solute/ volume of solution.

Mathematically:

Percent (mass/ volume) (%(m/v)) =

𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 (𝑔)

𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑜𝑓 𝑣𝑜𝑙𝑢𝑚𝑒 (𝑚𝐿)𝑥 100%

A solution contains 2.7 g CuSO4 in 75 mL of

solution. What is the percent (mass/ volume) of the

solution? Use % (mass/ volume) = m/ v x 100%:

%𝑚

𝑣=

2.7 𝑔 𝐶𝑢𝑆𝑂4

75 𝑚𝐿 𝑠𝑜𝑙𝑛𝑥 100 = 3.6% 𝑏𝑦 𝑚𝑎𝑠𝑠

Calculate the grams of solute required to make 250

mL of 0.10% MgSO4 (m/v).

Rearrange % (mass/ volume) = m/ v x 100%, to

solve for mass of solute:

𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 =(0.10%)(250𝑚𝐿)

100%= 0.25 𝑔 𝑀𝑔𝑆𝑂4

What is the concentration, in percent (m/ v), of a

solution with 75 g K2SO4 in 1500 mL of solution?

%𝑚

𝑣=

75 𝑔 𝑠𝑜𝑙𝑢𝑡𝑒

1500 𝑚𝐿𝑥 100 = 5.0%

Section 18.3

Obviously, the physical properties of a solution (salt water) differ from those of the pure solvent (water)

used to make the solution.

Some of these properties are the result only to the mere presence of solute particles in the solution, rather

than the identity of the solute particle itself.

These properties are called colligative properties.

Solutions have higher boiling points than the solvents that make them up.

We call this difference the boiling-point elevation.

When salt is dissolved in water for example, it takes additional kinetic energy for the solvent particles of water to overcome the attractive

forces that keep them in the liquid.

When a substance freezes, the particles of the solid take on an orderly pattern.

The presence of a solute (such as salt) in water disrupts the formation of this pattern, because of the shells of

water of solvation.

As a result, more kinetic energy must be withdrawn from a solution, than from its solvent, in order for it to

solidify.

The freezing point depression is the difference in temperature between the freezing point of a solution,

and that of its pure solvent.

Table 1. Experimental Freezing Point Depression a. Freezing Point of Water (Tf) 0°C b. Freezing Point of Solution (Tn) 1 pm: -6.07; 2 pm: -6.02°C c. Freezing Point Depression (∆T = Tn – Tf) ________°C

Table 2. Theoretical Freezing Point Depression a. Mass of solute (salt) used 1 pm: 800 – 310g = 490 g

2 pm: 800 – 350g = 450 g b. Moles of solute used = (_____g NaCl used)(1 mol NaCl/58.44 g NaCl)

________mol

c. Mass of solution 1 pm: 3,310 kg; 2 pm: 2,950 g d. Mass of solvent (water) ________g e. Molality (m) of solution (mol solute/ kg solvent) ________mol/ kg f. van’t Hoff Factor for NaCl (i) 2 g. Molal Freezing Point Depression Constant for H2O (Kf)

1.86° C/m

h. Theoretical Freezing Point Depression (∆T = Kf m i) ∆T = (1.86°C/m)(______m)(2) = ________°C

Table 3. Calculate Percent Error Experimental Freezing Point Depression (from Table 1[c]) ________°C Theoretical Freezing Point Depression (from Table 2[h]) ________°C Percent Error (% Error = Experimental/ Theoretical x 100) ________ % error

Section Review 18.3

24. Why does a solution have a lower vapor pressure

than the pure solvent of that solution?

The introduction of solute molecules reduces the number of solvent molecules with enough kinetic energy to escape.

Section Review 18.3

27. An equal number of moles of KI and MgF2 are

dissolved in equal volumes of water. Which solution

has the higher

a. boiling point?

b. vapor pressure?

c. freezing point?

MgF2(aq)

KI(aq) KI(aq)

You will be responsible for all calculations on solutions, in your final exam.

These calculations include: molarity (M), pp. 509 - 511 dilutions (M1V1 = M2V2) pp. 511, 512 percent compositions, pp. 513, 514 percent by volume percent by mass boiling point elevation freezing point depression, pp. 518, 519 molality (m), p. 520 do not worry about mole fraction - skip

Section 18.4

Colligative properties of a solution depend only on solute concentration, or the ratio of the number of solute particles to solvent particles. There are two convenient

ways of expressing this ratio:

1. Molality (m) – number of moles of solute dissolved per kilogram (1000 g) of solvent. It is also known as

the molal concentration (in contrast to molar concentration [moles solute/ liter soln]).

2. Mole Fraction – the ratio of the moles of solute in solution to the total number of moles of solvent and

solute.

Calculate the molality of the following solutions.

a. battery acid containing 368 g H2SO4 in 857 g

H2O

b. salt water containing 15.8 g NaCl in 150.g H2O

c. 30.0 g sucrose (C12H22O11) 70.0 g H2O (molar

mass of sucrose is 342.3 g/mol