Discussion No. 4

The Quantitative Expression of Concentration of Solutions

and Colloidal Dispersions

Marris Ellyanna Camille B. Lam, Jyvush Brandis C. Laya,

Arvy Llena, Arthur Odil T. Macatula

Section 1NUR-9 Group 4

INTRODUCTION

The concentration of a chemical solution describes the amount of solute that is

dissolved in a given amount of solvent. There exist several types of solutions: solids dis-

solved in liquids, solids dissolved in solids, to name a few. In these types of solution,

both solute and solvent exists either in the same or different phases. As a consequence,

there are also a number of ways to express the relative amounts of solute and solvent in

a solution: qualitatively, semi-qualitatively and quantitatively. In this discussion, we will

be studying how concentration of solutions and colloidal dispersions will be expressed

quantitatively.

GENERAL OBJECTIVES

At the end of the laboratory discussion, the students should be able to understand the

different ways by which concentration of solutions and colloidal dispersions are quanti-

tatively expressed

PERCENT CONCENTRATIONS

PERCENT CONCENTRATIONS

• Percent Concentration = ( Amount of Solute ) / ( Amount of Solution ) X

100%

PERCENT BY MASS

• The mass percent is used to express the concentration of a solution when the

mass of a solute and the mass of solution is given.

• (%m/m)

• Mass Percent = (Mass of Solute) / (Mass of Solution) X 100%

PERCENT BY VOLUME

• The volume percent is used to express the concentration of a solution when the

volume of a solute and the volume of a solution is given.

• (%v/v)

• Volume Percent = (Volume of Solute) / (Volume of Solution) X 100%

PERCENT WEIGH BY VOLUME

• Also known as Mass/Volume Percent.

• The Mass/Volume Percent is used to express the concentration of a solution

when the mass of the solute and the volume of solution is given.

• (%m/v)

• Mass/Volume Percent = (Mass of Solute) / (Volume of Solution) X 100%

EXAMPLES

• Percent Concentration

-An aqueous solution contains 0.011 g of sulfuric acid and 2,000 grams of

water.

• Percent by Mass/Volume

-What is the weight/volume percentage concentration of 250mL of aqueous

sodium chloride solution containing 5g NaCl?

    mass solute (NaCl) = 5g 

    volume of solution = 250mL

w/v (%) = 5g ÷ 250mL x 100 = 2g/100mL (%m/v)

• Percent by Volume

-You have 40.0ml of C6H5CH3 that is put into 75.0ml of C6H6, what is the

percent volume of C6H5CH3?

Volume of Solute = 40.0ml

Volume of Solution = 40.0ml+75.0ml = 115.0ml

(40.0ml)/(115.0ml) X 100 = 34.8 % of C6H5CH3

• Percent by Mass

-5.45 g of NaCl is added to 100 mL of water. What is the weight percent?

Mass of Solute = 5.45g

Mass of Solution = grams water + grams NaCl = 100g +5.45g = 105.45g

(5.45g) /(105.45g) X 100 = 5.17 (%m/m)

*Remember that: 100ml x 1.00g/ml = 100g

Osmolarity

Osmolarity - is defined as the number of osmoles of solute per liter (L) of solution.

It is expressed in terms of Osmol/L or Osm/L

For Example:

A 1 mol/L NaCl solution has an osmolarity of 2 osmol/L. A mole of NaCl separates fully

in water to yield two moles of particles: Na+ ions and Cl- ions. Each mole of NaCl be-

comes two osmoles in solution.

Osmoles - Osmolarity is defined in terms of osmoles. An osmole is a unit of measure-

ment that describes the number of moles of a compound that contribute to the osmotic

pressure of a chemical solution. It is related to osmosis and is used in reference to so-

lution where osmotic pressure is important, such as blood and urine.

Sample Problem

You have 400 mL of a 15 % MgCl2 solution

Molecular Weight: Mg = 24.3, Cl = 35.5)

Each mole of MgCl2= 3 osmoles in solution)

1. Since this is a 15% solution, how many grams of MgCl2 are in 100 mL of the so-

lution?

15% means 15g/100mL

= 15g

2. If there are 15 g/100 mL, then how many grams total are there in the 400 mL?

Equate

15 g/100 mL = x g/400 mL (Cross multiply)

15(400) = 100 x

x = 15(400) / 100

x = 60

= 60g

3. How many grams does 1 mole of MgCl2 weigh?

Substitute the molecular weights

MgCl2 = (24.3) + 2(35.5)

= 24.3 + 71

= 95.3 g

4. If there are 95.3 g in one mole, then how many moles are there in the 400 mL

MgCl2 solution?

There are 60 g in 400 mL, Convert Grams to Moles

(60 g) (1 mole/95.3 g) = 0.63 moles

5. How many osmoles are in the 400 mL MgCl2 solution?

Each mole is 3 osmoles. Convert moles to osmoles.

0.63 moles (3 osmole/1 mole) = 1.89 osmoles

6. What is the osmolarity of the MgCl2 solution?

With 1.89 osmoles in 400 ml, how many osmoles would be in 1 liter?

Convert Mililiter to Liter

1.89 osmoles/400 ml (1000ml/l) = 4.73 osmoles/liter

Osmolarity is 4.73 Osm/L

Dilution

Dilution is the addition of solvent to decrease the concentration of the solution. Exam-

ple: If a glass of orange juice is too concentrated to drink, just simply add enough water

to dilute the juice until it is concentrated enough to pleasantly drink.

-Keep in mind that the amount of solute is constant. Only the volume of the solution is

changed by adding solvent.

-Law of Conservation of Mass is applied when calculating dilution.

-Law of COnservation of Mass: In a chemical reaction, the mass of the products equals

the mass of the reactants.

Formula

M1V1 = M2V2

M1V1 = Initial Value

M2V2 = Final Value

M = Concentration (molarity)

V = Volume (Liters, milliliter)

Example

What is the concentration of a solution formed by diluting 5.0 mL of a 3.2 M glucose so-

lution to 40.0 mL?

M1 = 3.2 M V1 = 5.o mL

M2 = ? V2 = 40.0 mL

3.2 M x 5.0 mL = M2 x 40.0 mL

M2 = (3.2 M x 5.0 mL)/ 40.0 mL

M2 = 0.40

How many milliliters of 5.5 M NaOH are needed to prepare 300 mL of 1.2 M NaOH?

←M1 = 5.5 M V1 = ?

←M2 = 1.2 M V2 = 300 mL

5.5 M x V1 = 1.2 M x 300 mL

V1 = (1.2 M x 300 mL)/ 5.5 M

V1 = 65 mL

MOLARITY AND MOLALITY

• Molarity - It is defined as the number of moles of solute dissolved in 1L of solu-

tion.

FORMULA:

PROBLEMS:

1. Find the molarity of a solution that contains 49.3 grams of NaCl in enough water

to make 500.0mL of solution.

49.3 x 1/58.5g = 0.843 mol NaCl

M = 0.843 mol NaCl

0.5 L

= 1.69 M

2. Calculate the molarity of a solution prepared by dissolving 23.7 grams of

KMnO4 into enough water to make 750 mL of solution.

23.7 x 1mol/158g = 0.15mol

M = 0.15mol

0.750L

= 0.2 M

3. Find the molarity of a solution that contains 93.7 grams of KNO3 in enough to

make 750cm³ of solution.

93.7g x 1mol/101.1g = 0.977mol

M = 0.977mol

0.75L

= 3.71 M

4. What is the molarity of the solution if 36.9 grams of CaCl2 is dissolved in 750mL

of an aqueous solution? The density of the solutions = 1.21g/mL

Solution = Solute + Solvent

908g – 36.9 = 871.1g

36.9 x 1mol/101.1 = 0.332mol

M = 0.332mol

0.75mL

= 0.443M

· Molality - It is defined as the number of moles of solute dissolved in 1kg (1000g)

of solvent.

FORMULA:

PROBLEMS:

1. A 4 g sugar cube (Sucrose: C12H22O11) is dissolved in a 350 ml teacup of 80 °C

water. What is the molality of the sugar solution? Given: Density of water at 80° =

0.975 g/ml

4g x 1mol/342g = 0.0117mol

Mass = (0.975 g/mL)(350mL) = 341.25g

Mass = 0.341kg

m = mol/kg

= 0.0117mol/0.341kg

= 0.034mol/kg

2. What is the molality of a solution which 49 grams of H2SO4 is dissolved in 250g

of water?

49g x 1mol/98g = 0.5mol

m = 0.5mol/0.5kg

= 2mol/kg

3. What is the molarity of the solution if 36.9 grams of CaCl2 is dissolved in 750mL

of an aqueous solution? The density of the solutions = 1.21g/mL

Solution = Solute + Solvent

908g – 36.9 = 871.1g

36.9 x 1mol/101.1 = 0.332mol

m = 0.332mol/0.8711kg

= 0.381mol/kg

4. What is the molality of the solution if 14.7 of NaCl is dissolved in 100g of water?

The density of the solution = 1.15g/mL

14.7g + 100g = 114.7g

m = mol/kg

= 0.251mol/0.1kg

= 2.51 mol/kg