Grade 12 GENERAL CHEMISTRY 2
Module 2
Wk2
Objectives:
At the end of the lesson the students must be able to:
1. Interpret the phase diagram of water and carbon dioxide.
2.Determine and explain the heating and cooling curve of a substances.
3 Use the different ways of expressing the concentration of solutions: percent by mass, percent
by volume , percent by volume-mass, mole fraction, molarity, molality parts per million (ppm)
4.Perform stoichiometric calculations for reaction in solutions
5.Describe the effect of concentration on colligative properties of solutions
Introduction:
Solutions were produced which resulted in only one phase and the components such as solute
which is the substance to be dissolved , less in amount and solvent which is the dissolving medium and
greater in amount,.and the types are solids, liquids, and gas. This is one of the basis wherein a solution
being classified as dilute and concentrated.It is impossible to identify the concentration by mere of
taste, color, or any but rather quantitatively measured such as mole fraction, molarity, molality, percent
by mass and volume.
Colligative properties is also the basis of concentration of solute whether there is an increase in
the boiling point, lowering of the freezing point and vapor pressure lowering or osmosis which is
phenomenon of solve flow through a semipermeable to equalize the solute concentration on both
sides of the membrane.
The change in phase from solid, liquid and gas resulted to the processes of evaporation, freezing,
melting,condensation , deposition and sublimation are expected to produce considering the effect of
temperature.
Discussions:
Concentration of Solutions:
The amount or concentration of solute in a given amount in a given amount of solvent can be
expressed in several ways. If a very small amount of sugar is dissolved in large amount of water. .
More accurate way of expressing the concentration could be expressed quantitatively.
A. Percentage (%) Mass:
Percentage concentration by Mass: The ratio of the mass of solute ÷ mass of solution .
(%)Mass ratio is a decimal fraction that represents the mass in grams of 1 gram of
solution.Therefore , 100 times the mass ratio is the mass in grams of solute in 100 g - grams of
solute per 100 g of solution .
% concentration by mass = mass of solute x100
Mass of solution
Examples;
1, Find the concentration of a solution made by dissolving 50 g of alcohol in enough water to make
100 g of solution .
Given:
Mass of solute = 50 g
Mass of solution = 100 g
Solution:
Percent by Mass ( %) = mass of solute
% by mass
-------------------x100
Mass of solution
= 50 g
_____x100 =50 %
100g
2. How many grams of NaCl are present in 250 g salt solution containing 9% NaCl ?
Given: mass of salt solution = 250 g
% by mass of solution = 9 %
Solution :
mass of NaCl = mass of solution x % concentration of solution
_________________________________________
100
Mass of NaCl = 9% x 250 g =22.5 g
__________
100
Activity # 1 Solve the following problem. Specify the correct units and observe the number of
significant figure.
1. When 125 g of a solution was evaporated to dryness , 42.3 g of solute was recovered. What was
the percentage concentration of the solution by mass?
2. How many grams of glucose C6 H12 O6 and of water are in 500 g of 5.3 % by mass of glucose of
solution?
B. Percent by Volume:Divide the volume of solute by the volume of solution then multiplied by
100.
% by volume = volume of solute
______________x 100
Volume of solution
Examples:
1.Supposed we mixed 50 mL of alcohol with enough water to make 400 mL of solution
.Find the percentage by volume of the solution .
Given : volume of alcohol (solute ) = 50 mL
Volume of solution
= 400 mL.
Volume of solution = volume of solute + volume solvent
Solution % by volume = 50 mL:
_________x100 = 12.5%
400m L
2.Rubbing alcohol in aqueous solution contains 70% isopropyl alcohol (C 3H7OH) by volume . How
would you prepare 250 mL rubbing alcohol from pure isopropyl alcohol ?
Volume of isopropyl alcohol = % by volume of solute x100
_________________
Volume of solution
Volume of solute = volume of rubbing alcohol x % by volume
_______________________________
100
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Volume of solute = 70% x 250 mL
_______________175 mL
_____
100
Activity # 2
1, Find the percentage of a solution that is made by mixing 150 mL OF sodium silicate
(Na2SiO4 ) in 800 mL solution.
2 How much ammonium sulfate is required to prepare 400 mL in 75 % solution ?
C. Percent by Weight - Volume = mass of solute in grams x 100
_______________________
Volume of solution
Examples :
1. If 30 g of ethyl alcohol is mixed with water to make 100 mL solution , what would be the
% by mass - volume of alcohol?
Given :mass of solute (ethyl alcohol ) = 30 g
Volume of solution = 100 mL
Solution:% by mass- volume = mass of alcohol ( solute ) x100
____________________
Volume of solution
= 30g x100
_____________x 100 =30%
100 mL
2. Find the percent by mass - volume of a solution that is made by mixing 2.0 g of iodine in
enough carbon tetra chloride to make 80 mL solution.
% mass -volume = 30 g
_________ x100
80 mL
= 2.5 %
Activity # 3
1. What is the % by mass- volume of a solution made by dissolving 610 g of ammoniium
chloride in enough water to prepare 1200 mL of solution ?
2. Find the mass -volume by mixing 400 g of salt in water to yield 800 ml of solution..
D. Parts Per Million (ppm ) One parts per million is equivalent to 1 mg of solute per liter of solution.
1 ppm= 1mg of solute / 1 l of solution
Example What is the parts per million of 30 g of NaCl and 105 g of water?
Solution:
Mass of solution = mass of solvent + mass of solvent: 30g NaCl + 105 g water= 135 g of
solution
Ppm = 30g / 135 = .222 x 1000000 = 222,
Activity # 3
1. What is the ppm of 40 g sugar in 100 g of water?
2. It was reported that canned evaporated milk con tained up to 3.2 ppm of lead. At this
concentration , How many grams of lead are present in 8 ounce (470 mL) of evaporated milk.
E. Mole Fraction (X) =is the ratio of the number of moles of a given component to the total
number of moles of solution.
Mole fraction (X) of solute (A) = moles of solute
X A = mole fraction of A ( Solute)
XB= Mole fraction of B (solvent)
The graph shows that melting and vaporization require heat absorptio of heat. An increase in
temperature as the substances changes from solid to liquid to gas. Starting with a solid as heat
is added , the temperature also increases,.. In melting , the temperature remain constant and
this is the temperature , where the solid is being converted to liquid . Both solid and liquid
coexist. When all the solid has been converted to the liquid state, temperature again rises as
heat is continously added, When the liquid is being converted to its vapor, the temperature
again remains constant it is boiling point,\. When all of the liquid has been converted to vapor,
temperature rises again.
Mole fraction (X) mole of solute ( B ) =moles of solvent
X B= moles of solvent
____________
Moles of solution
Mole fraction of solute +mole fraction of solvent
XA + XB = 1
Example;
A solution is made by dissolving 1.25 g Na2 SO4 in 65 g of water . Calculate the mole
fraction of solute and the solvent
Given 1.25 g Na2 SO4
Mass of solvent (water ) = 65 g
Step 1. Determine the molar mass and the atomic mass
Atomic mass from Periodical Table :
Na= 23 g / mol
O= 16 g / mol S= 32 g /mol H = 1 g/mol
Molar mass: Na2SO4
H2O
Na= 2x 23= 46 g /mol
H= 2x 1= 2 g/ mol
S = 1x 32= 32 g/ mol
O = 1x16= 16 g / mol
O = 4x 16 = 64 g/ mol
+
---18 g / mol
______142 g / mol
Step 2. Determine the number of moles of solute and solvent :
Moles of Na2SO4 and H2O
Mole of Na2 SO4 ( nA) = 1,25 g / 142 g/mol = .008 mol
Mole of H2O ( nB ) =65 g / 18 g/ mol = 3.16 mol
Step 3: Determine the total moles
Total moles ( nt) = n A + n B = .008mole + 3. 6 mol = 3.618 = 3. 62 mol
Step 4 .Determine the mole fraction of solute + mole fraction of solvent
XA .0083 moles / 3.62mole2 = .0022
XB= 3. 62moles / 3. 62mol;es = 1
Xt = XA + XB: .0083 + 1= 1. 0083
Activity # 4
Determine the mole fraction of the following problems:
1. A solution is prepared by mixing 1 g of ethyl alcohol (C2H5OH) with 100 g of water . Calculate
the mole fractions of ethyl alcohol and water.
2. 36 g of H2O and 46 g of glycerin C3H5 (OH) 3
F Molarity -m the number of moles of solute in a liter of a solution :
Molarity (M)= moles of solute
_______________
Volume in a liter of solution
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Sample problem: What is the molarity (M) of 0.5 g NaOH present 2 L of the solution?
Solution: Atomic weight (Periodic Table) N6 g /mol = 23 g /mol O =16g / mol
H= 1g / mol
Molar mass of NaOH: Na=1x23 g/ mol = 23 g / mol
O = 1 x 16 g / mol= 16 g / mol
H = 1x 1 g / mol = 1g/mol
Molar mass = 40 g /mol
Moles of solute = 0.5 g / mol / 40 g / mol= .0125 mol
Molarity (M) = moles of solute / Volume in litre of solution
M = .00125 mole / 2l = .000625 mol / liter
2What is the weight of H2SO4 is present in .2 M in 500 mL solution ?
Solution : Atomic weights (periodic table ) H= 1 g /mol S= 32 g / mol O= 16 g /mol
Molar mass:
H = 2x 1 g /mol
conversion factor : I L = 500 mL
S= 1x 32 g / mol
O= 4 x 12 g / mol
__________________
98 g /mol
Mass of solute = M x molar mass x V (L) = ..2mol / L x 98 g / mol x .500L = 9.8 g
Activity #5
1. Calculate the M of a solution containing 40 g NaOH in 200 mL’
2. What is the molar concentration f of a solution 16 g of methanol CH3OH IN 200 mL solution
?
G. Molality is a measure of number of moles of solute present in 1 kg of solvent.
FORMULA
Molality (m)= moles of solute / kg of solvent
Conversion : 1kg = 1000g .
Note: Molarity is written in capital M while molality is m. this should be followed strictly in expressing
these two concentrations .
Example: What is the molality of a solution in which 20 g Ca(OH) 2 (molecular weight = 74 g/
mol) dissolved in 250 g of water ? solution : 1000g = 1 kg
Conversion : 250 g x 1kg / 1000 =.250 kg
Moles of solute = 20g / .74 g/mol = .270 mol
Molality (m) = .270 mol / .250 kg= 1.08 mol / kg
Activity # 6
1. What is the molality of a solution which cotains 20 g o cane sugar C 12H22O11 dissolved in 125 g
of water ?
2. Calculate the mass of MgCl2 which is 0.25 m in in 250 grams of water ?
Stoichiometry is the field of chemistry that is concerned with the relative quantities of reactants
and products in chemical reactions. For any balanced chemical reaction, whole numbers
(coefficients) are used to show the quantities (generally in moles ) of both the reactants and
products.
Stoichiometric Calculations for Reactions in Solution:
Rules to follow in calculating reactions in solutions;
Balance the equation.
 Convert units of a given substance to moles.
 Using the mole ratio, calculate the moles of substance yielded by the reaction.
 Convert moles of wanted substance to desired units
.Relating Masses of Reactants and Products
What mass of sodium hydroxide, NaOH, would be required to produce 16 g of the antacid
milk of magnesia [magnesium hydroxide, Mg(OH)2] by the following reaction?
MgCl2(aq)+2NaOH(aq)⟶Mg(OH)2(s)+NaCl(aq)MgCl2(aq)+2NaOH(aq)⟶Mg(OH)2(s)+Nl(aq
)
Solution
Determine the molar masses
NaOH : Na= 1x23 =23g /mol O= 1x16 =16 g/mol H x 1=1 g/mol then add = 40g/ mol
Mg(OH)2 : Mg= 24x 1=24 g/mol O = 2 x 16=32 g/mol H 2x1= 2 g/mol = 58 g/mol
Ratio based to the balanced equation 2mol of NaOH in 1 mol of Mg(OH)2
Solution:
16 g Mg(OH)2 x 1mol Mg(OH)2 x 2 mol NaOH X 40 g NaOH
_____________ ____________
------------- ==22 g NaOH
58 g of Mg(OH)2 1 mol Mg(OH) 1 mol NaOH
Relating Masses of Reactants
What mass of oxygen gas, O2, from the air is consumed in the combustion of 702 g of
octane, C8H18, one of the principal components of gasoline?
2C8H18+25O2⟶16CO2+18H2O
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.2
Water has an unusual phase diagram: its melting point decreases with increasing pressure
because ice is less dense than liquid water. The phase diagram of carbon dioxide shows
that liquid carbon dioxide cannot exist at atmospheric pressure. Consequently, solid carbon
dioxide sublimes directly to a gas.
Unlike carbon dioxide and most other substances, the phase diagram of water shows a
negative slope for the boundary line between the liquid and solid state. This difference has
to do with that fact that water actually expands as it goes from the liquid state to the soli
Both phase diagrams for water and carbon dioxide have the same general Y-shape, just
shifted relative to one another. This shift occurs because the liquid phase in the dry
ice can only occur at higher temperatures and pressures, whereas, in ice the
liquid phase occurs at lower temperatures and pressures.
Molar mass C8H18 = 114.23 g/mol O2 = 32 g/ mol
Ratio : 25 mol O2 : : 2 mol of C8 H18
Solution
702g C8H18 x 1mol C8H18 x 25 mol O2 x 32 g O2
_________: __________ _______ =2.46 x 10 3 g of Oxygen
114.3 g
2 mol C8 H18 mol O2
Activity:
1. What mass of CO is required to react with 25.13 g of Fe2O3 ?
Activity# 8
Based from the graph interpret how do three phases of matter formed in water and carbon
dioxide.
Fe2O3+CO⟶Fe+CO2 ?
2. What mass of gallium oxide, Ga2O3, can be prepared from 29.0 g of gallium
metal?
4Ga+3O2⟶2Ga2O3.
Colligative Properties. Colligative properties of solutions are properties that depend upon
the concentration of solute molecules or ions, but not upon the identity of the solute. Colligative
properties include vapor pressure lowering, boiling point elevation, freezing point depression, and
osmotic pressure.
The colligative properties of a solution differ from that pure solvent at it depends only on the
number of particles of solute present but not on the number of particles of solute present but not on
the kind and nature of solute. Since non electrolyte solute do not ionize in solution ,the number of
solute particles will be less in a solution containing non electrolyte solute .
The vapor pressure of a solution is lower than the vapor pressure of a pure solvent to escape. If
the escaping tendency of the molecule, then its vapor pressure becomes lower
Osmosis - allows the passage of certain molecules only through the semipermeable membrane
membrane . If two solutions are placed adjacent each other , the movement of the solvent is from a
region of lower concentration. In a region of higher concentration to a lower concentration until the
concentration of both of the solution are equal.
Boiling point Elevation the addition of solute to a solvent causes its boiling point to increase while
in freezing point addition of solute causes freezing point of its solution to lower
Activity # 7 Explain the following problems.
1. Which has a lower freezing point, pure water or water urea mixture? Why?
2. Why do ice cream vendor sprinkled salt to their ice cream ?
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The graph shows that melting and vaporization require heat absorptio of heat. An increase in
temperature as the substances changes from solid to liquid to gas. Starting with a solid as heat
is added , the temperature also increases,.. In melting , the temperature remain constant and
this is the temperature , where the solid is being converted to liquid . Both solid and liquid
coexist. When all the solid has been converted to the liquid state, temperature again rises as
heat is continously added, When the liquid is being converted to its vapor, the temperature
again remains constant it is boiling point, When all of the liquid has been converted to vapor,
temperature rises again
Heating Curve Diagram
Activity 38 Based from the graph . answer the following questions . Determine the phase
change in the processes.
A freezing
B. Melting
C. Vaporization
D. Condensation .
Agreement cite an example where freezing depression is observed in our environment.
References; Chemistry : 8th edition Raymond Chang
General Chemistry : Cengage
Fundamentals of Chemistry Freidrick Redmore
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