SCHEME OF WORK
SCHEME OF WORK 2ND TERM
WEEK TOPIC
1 Revision / Introduction to the mole concept & Avogadro's number
2 The mole concept in terms of masses numbers, volumes, reactants and products.
3-4 Writing and Balancing of chemical equations. Calculations of:
i. Molar mass of the compound
ii. Percentage of an element in a compound
iii. Empirical and molecular formulae
5 Chemical laws and their verification
i. Law of conservation of mass
ii. Law of constant composition / definite proportions
iii. Law of multiple proportions
6-7 Chemical Combinations
i. Electrovalent bonding, properties of electrovalent compounds
ii. Covalent bonding, properties of covalent compounds
iii. Other types of bonding including : Co ordinate covalent bond (dative), hydrogen bond, metallic bond, Van-der-Waal's weak forces of attraction.
8-9 The kinetic theory of matter and the gas laws
i. Boyle's law
ii. Charles' law
iii. General gas equation
iv. Dalton's law of partial pressure
v. Ideal gas equation
10 Avogadro's law, Gay-Lussac's law of combining volumes, Graham's law of diffusion.
11 Revision
NEW
WEEK TOPIC
1. Revision of last terms work
2&3. Gas laws
4,5&6. Standard Separation Techniques for mixtures
7&8. Revision.
9&10. Examination.
REFERENCE TEXTS:
1 .Comprehensive Certificate Chemistry for Senior Secondary Schools by G N C Ohia et al
2. New School Chemistry for Senior Secondary Schools by Osei Yaw Ababio
3. Chemistry for Senior Secondary Schools 1 by Magbagbeola O, et al; Melrose Books and Publishers
4. Revised edition understanding chemistry for schools and colleges by Godwin O. Ojokuku.
2ND TERM
WEEK 1
TOPIC: GAS LAWS
CONTENTS
I. Boyle’s law and Charles’ law
II.General gas equation.
III. Gay-Lussac’s law and Avogadro’s law
SUB-TOPIC 1 BOYLES LAW
The relationship between volume and pressure of a gas was first started by Robert Boyle in 1662.
Boyle’s law states thatthe volume of a given mass of gas is inversely proportional to its pressure, provided that the temperature remain constant.
According to Boyle’s law, volume of a gas increases as the pressure decreases and vice versa.
This relationship is independent of the nature of the gas and it can be expressed mathematically as:
V ∝1/(P )
∴ V = k/p
Or PV =K
Where V= volume at pressure P
K= (a mathematical constant)
For a given mass of a gas, the product of its pressure and its volume is always a constant. If the pressure of a given mass of gas increases, its volume will decrease by a similar proportion and vice versa, as long as the temperature remain constant. This relationship can also be expressed mathematically as:
P1V1 = P2 V2
Where V1 = volume at pressure P1
V2= Volume at pressure P2
Boyle’s law can still be re-stated as: ‘The pressure of a given mass of gas is inversely proportional to its volume, provided the temperature remains constant.Boyle’s law can further be illustrated with the diagram below, showing that when P is increasing, V is decreasing and when P is decreasing, V is increasing.
P3
V3
P2
V2
P1
V1
P1 = Initial pressure
P2= Final pressure
V1=Initial volume
V2= final volume
Graphical representation of Boyle’s law
(c)
V P
P O 1/v
Worked examples
375cm3 of a gas has a pressure of 770mmHg. Find its volume if the pressure e is reduced to 750mmHg.
P1V1 = P2V2(Boyle’s law)
P1 = 770mmHg
P2 = 750mmHg
V1=375cm3
V2 =? (newvolume of gas)
P1V1 = P2 V2
V2 = (P1 V1)/P2 =(770 ×375)/750 =385cm3
The new volume will be 385cm3
100cm3 of a gas has pressure of 1 atmosphere. Determine the volume of the gas at 5 atmospheres keeping the temperature constant.
Solution: since T is constant, we are to use Boyle’s law.
P1→ Initial pressure = l atmosphere
P1→Final pressure= 5
V1→ Initial volume= 100cm3
V2 → (required quantity)
Recall: V2 = P1V1 = P2V2
V2=(P1 V1)/P2 =(100 ×1)/5 = 20cm3
CHARLES LAW
The effect of temperature changes on the volume of a given mass of gas at a constant pressure is described by Charles law.Charles law states thatthe volume of a given mass of gas is directly proportional to its temperature in Kelvin, provided that pressure remains constant.
The volume of the gas decreases as the temperature decreases, and increases as the temperature increases.
Mathematically, the law can be expressed as:
V∝ T
∴ V = KT
Or V/T = K
Where v= volume
T= Kelvin Temperature
K= mathematical constant
A Representation of Charles’s law
V1 V2 V3 V4
T1 T2 T3 T4
For a direct relationship, when the temperature increases, the volume will also increase at the same rate and vice versa, at constant pressure.The diagram above shows that when V is decreasing, T is also decreasing and when V is increasing, T is also increasing thus, making the quotient constant.
Charles’s law can be represented graphically has shown below.
V(m)3
273 0 Temperature 0C
V
V/1
T(k)T(k)
If we divide the varying gas volumes by the corresponding temperature in Kelvin, the result would always be a constant. This relationship can also be expressed in another form.
V1/T1 = V2/T2∴ V2=(T2 V1)/T1
Where V1 is the volume at temperature T1
V2 is the volume at temperature T2
TEMPERATURE CONVERSION
To convert from Celsius scale to Kelvin scale, add 273 i.e. T= 0C + 273. This is because O0C=273K.
To convert from Kelvin scale to Celsius scale, subtract 273. i.e
0C= T- 273.
Where T= Temperature in Kevin
0C= Temperature in Celsius.
Examples:
1.Convert the following Celsius temperature to Kelvin temperature.
1000C (b) 00C (c) -570C
Solution
Recall: T= 0C + 273
1000C= (100 + 273) = 373k
0C=(0 + 273) = (0 + 273) = 373k
-570c = (- 57 + 273)k = (273-57)= 216k
2.Convert the following Kelvin temperatures to Celsius temperature.
298k (b) 405k (b) 285k (d) 0k
Solution
Recall 00c = k – 273
298k = (298 – 273)0C= 250C
405k = (405 – 273)0C = 120C
0k = (0 – 273)0C = - 2730C
Worked examples on Charles’s law
A gas occupies a volume of 20.0dm3 at 373k. Its volume at 746k at that pressure will be?
Here pressure is constant. Charles’s law will applied.
V1=20.0dm3
T1 = 273k
T2= 746
V2= ?
Recall Charles’s law = V1/T1 =V2/T2V2=( V1 T2)/T1V2 = (20 ×746)/273 = 40.0dm3
EVALUATION:
State Boyle’s law
State Charles’s law
Express the two laws mathematically
Draw two different graphs showing Boyle’s andCharles’s laws.
CONTENTS
I. Boyle’s law and Charles’ law
II.General gas equation.
III. Gay-Lussac’s law and Avogadro’s law
SUB-TOPIC 1 BOYLES LAW
The relationship between volume and pressure of a gas was first started by Robert Boyle in 1662.
Boyle’s law states thatthe volume of a given mass of gas is inversely proportional to its pressure, provided that the temperature remain constant.
According to Boyle’s law, volume of a gas increases as the pressure decreases and vice versa.
This relationship is independent of the nature of the gas and it can be expressed mathematically as:
V ∝1/(P )
∴ V = k/p
Or PV =K
Where V= volume at pressure P
K= (a mathematical constant)
For a given mass of a gas, the product of its pressure and its volume is always a constant. If the pressure of a given mass of gas increases, its volume will decrease by a similar proportion and vice versa, as long as the temperature remain constant. This relationship can also be expressed mathematically as:
P1V1 = P2 V2
Where V1 = volume at pressure P1
V2= Volume at pressure P2
Boyle’s law can still be re-stated as: ‘The pressure of a given mass of gas is inversely proportional to its volume, provided the temperature remains constant.Boyle’s law can further be illustrated with the diagram below, showing that when P is increasing, V is decreasing and when P is decreasing, V is increasing.
P3
V3
P2
V2
P1
V1
P1 = Initial pressure
P2= Final pressure
V1=Initial volume
V2= final volume
Graphical representation of Boyle’s law
(c)
V P
P O 1/v
Worked examples
375cm3 of a gas has a pressure of 770mmHg. Find its volume if the pressure e is reduced to 750mmHg.
P1V1 = P2V2(Boyle’s law)
P1 = 770mmHg
P2 = 750mmHg
V1=375cm3
V2 =? (newvolume of gas)
P1V1 = P2 V2
V2 = (P1 V1)/P2 =(770 ×375)/750 =385cm3
The new volume will be 385cm3
100cm3 of a gas has pressure of 1 atmosphere. Determine the volume of the gas at 5 atmospheres keeping the temperature constant.
Solution: since T is constant, we are to use Boyle’s law.
P1→ Initial pressure = l atmosphere
P1→Final pressure= 5
V1→ Initial volume= 100cm3
V2 → (required quantity)
Recall: V2 = P1V1 = P2V2
V2=(P1 V1)/P2 =(100 ×1)/5 = 20cm3
CHARLES LAW
The effect of temperature changes on the volume of a given mass of gas at a constant pressure is described by Charles law.Charles law states thatthe volume of a given mass of gas is directly proportional to its temperature in Kelvin, provided that pressure remains constant.
The volume of the gas decreases as the temperature decreases, and increases as the temperature increases.
Mathematically, the law can be expressed as:
V∝ T
∴ V = KT
Or V/T = K
Where v= volume
T= Kelvin Temperature
K= mathematical constant
A Representation of Charles’s law
V1 V2 V3 V4
T1 T2 T3 T4
For a direct relationship, when the temperature increases, the volume will also increase at the same rate and vice versa, at constant pressure.The diagram above shows that when V is decreasing, T is also decreasing and when V is increasing, T is also increasing thus, making the quotient constant.
Charles’s law can be represented graphically has shown below.
V(m)3
273 0 Temperature 0C
V
V/1
T(k)T(k)
If we divide the varying gas volumes by the corresponding temperature in Kelvin, the result would always be a constant. This relationship can also be expressed in another form.
V1/T1 = V2/T2∴ V2=(T2 V1)/T1
Where V1 is the volume at temperature T1
V2 is the volume at temperature T2
TEMPERATURE CONVERSION
To convert from Celsius scale to Kelvin scale, add 273 i.e. T= 0C + 273. This is because O0C=273K.
To convert from Kelvin scale to Celsius scale, subtract 273. i.e
0C= T- 273.
Where T= Temperature in Kevin
0C= Temperature in Celsius.
Examples:
1.Convert the following Celsius temperature to Kelvin temperature.
1000C (b) 00C (c) -570C
Solution
Recall: T= 0C + 273
1000C= (100 + 273) = 373k
0C=(0 + 273) = (0 + 273) = 373k
-570c = (- 57 + 273)k = (273-57)= 216k
2.Convert the following Kelvin temperatures to Celsius temperature.
298k (b) 405k (b) 285k (d) 0k
Solution
Recall 00c = k – 273
298k = (298 – 273)0C= 250C
405k = (405 – 273)0C = 120C
0k = (0 – 273)0C = - 2730C
Worked examples on Charles’s law
A gas occupies a volume of 20.0dm3 at 373k. Its volume at 746k at that pressure will be?
Here pressure is constant. Charles’s law will applied.
V1=20.0dm3
T1 = 273k
T2= 746
V2= ?
Recall Charles’s law = V1/T1 =V2/T2V2=( V1 T2)/T1V2 = (20 ×746)/273 = 40.0dm3
EVALUATION:
State Boyle’s law
State Charles’s law
Express the two laws mathematically
Draw two different graphs showing Boyle’s andCharles’s laws.
WEEK 2
SUB-TOPIC 2: GENERAL GAS EQUATION
From the gas laws, we know that the volume of a gas depends on both its temperature and pressure. The relationship between the three variable; i.e. volume,, temperature and pressure can be summarized up as follows:
If V ∝1/(P ) (Boyle’s law at constant temperature) and V ∝ T (Charle’s law at constant pressure)
V ∝1/(P ) × T (both temperature and pressure may vary) orPV/T = K (a mathematical constant for a fixed mass of gas)
PV/T =K is often known as the general gas equation.
GENERAL GAS EQUATION
General gas equation states that for fixed mass of a gas under any set of conditions of V, P and T, the value of PV/T must remain constant. If for a fixed mass of gas V1 is the volume at pressure
P1 and absolute temperature T1 and V2 is the volume at pressure P2 and absolute temperature T2 it follows that.
P1V1/T =P2V2/T2
The general gas equation can be used to find the volume of a gas when both its pressure and temperature change. Thus;
V2 =( P1 V1 T2)/(P2 V2)
The standard temperature and pressure
The value of gases are sometimes given in standard temperature and pressure (S. T. P). These values are standard temperature= 273k and standard pressure = 760mmHg. The S.I unit of standard pressure when used is 1.01 × 103Nm-2
Examples
At S. T. P a certain mass of gas occupies a volume of 790cm3, find the temperature at which the gas occupies 1000cm3 and has a presence of 720mmHg
P1V1/T1= P2V2/T2
P1 = 760mmHg (at stp), V1= 790cm3
T1 = 273k (at stp), V2 = 1000cm3
P2 = 726mmHg
T2 = New Temperature
∴T2 =(P2 V2 T1)/(P1 V1)
= (720 ×1000 ×273)/(760 ×790) = 330.1k
The new temperature of the gas is 330.1k
A given mass of gas occupies 850cm3at 320k and 0.92 × 103Nm-2
of pressure. Calculate the volume of the gas at S.T.P.
P1V1/T1 = P2V2/T2
P1= 0.92 × 103Nm-2 T1= 320k
V1= 850cm3 P2= SP + 1.01 × 103Nm-2
T2= 273k (at stp)
V2 = new volume of gas.
∴V2 = (P1 V1 T2)/(P2 T1) = (0.92 ×850 ×273)/(1.01 ×〖10〗^3 ×320) = 660.5〖cm〗^3
EVALUATION
Explain the general gas equation.
If the volume of a given mass of a gas at 298k and a pressure of 205.2 ×〖10〗^5 〖Nm〗^(-2) is 2.12〖dm〗^3. what is the volume of the gas S.T.P (standard pressure= 101.3 ×103Nm-2, standard temperature = 273)
SUB TOPIC 3;GAY- LUSSAC’S LAW AND AVOGADRO’S LAW
Gay- Lussac’s law describes the combining volumes of gases that react together.In his experiment, all temperatures and pressures were kept constant:
A. STEAM: Gay- Lussac’s observed that two volumes of hydrogen reacted with one volume of oxygen to yield two volumes of steam
Hydrogen + Oxygen Steam
Volume 2 : 1 2
Ratio 2 : 1 2
B. HYDROGEN CHLORIDE GAS: One volume of hydrogen combined with one volume of chloride to yield two volumes of hydrogen.
Hydrogen + Chlorine Hydrogen Chloride
i.e. H2 + CL2 2HCL
Volume= 1 : 1 2
Ratio = 1 : 1 2
C.Carbon monoxide + Oxygen + CarbonIVoxide
2CO + O2 2CO2
Ratio 2 : 1 : 2
Gay- Lussac’s noticed that the combining volumes as well as the volumes of the products, if gaseous, were related by simple ratios of whole numbers. He proposed the law of combining volume or gaseous volumes.
Hence; Gay- Lussac’s law combining volumes states thatwhen gases react, they do so in volumes which are simple ratios to one another and to the volumes of the producers, if gaseous provide that the temperature and the pressure remain constant.
EXAMPLES
1.What is the volume of oxygen required to burn completely 45〖cm〗^3 of methane (〖CH〗_4)?
Equation of reaction:
〖 CH〗_4(g) + 〖 20〗_(2(g))) 〖Co 〗_(2(g)) + 2〖HO〗_(2(g))
Vol: 1 2 1 2
Ratio:1 2 1 2
By Gay- Lussac’s Law:
1 volume of methane required 2 volumes of oxygen i.e.
1〖cm〗^3 of methane requires 2〖cm〗^3 of oxygen
∴ 45〖cm〗^3 of methane require 90〖cm〗^3 of oxygen
3.20〖cm〗^3 of carbonIIoxide are sparked with 20〖cm〗^3 of oxygen. If all the volumes of gases are measured at a S.T.P, calculate the volume of the residual gases after sparking?
Equation of reaction 2〖Co〗_g+ O_2g 2〖Co〗_2g
Combining volume 2 : 1 : 2
Volumes before sparking 20〖cm〗^3 10〖cm〗^3, 20〖cm〗^3
Volumes after sparking -10 20
Residual gases = un-reacted oxygen + carbon (IV) oxide formed
Volume of residual gas = 10〖cm〗^3+ 20〖cm〗^3= 30〖cm〗^3
AVOGADRO’S LAW
Avogadro’s Law states thatequal volumes of all gases at the same temperature and pressure contain the same number of molecules.
This law means that for any type of gas,e.g. oxygen hydrogen, Chlorine etc if their volumes are the same, they will have the same number of molecules.
Avogadro’s Law is easily applied to convert volume of gases to the number of molecules. Avogadro’s Law can be used to solve problem under Gay –Lussac’s law of combining volumes.
The formation of steam from reaction of Hydrogen and Oxygen is given below:
Reaction Hydrogen + Oxygen Steam
Volume: 2 1 2
Gay –Lussac’s: 2 : 1 : 2
Avogadro’s Law: 2 : 1 : 2
This agrees with the equation below:
2H_(2(g)) + O_(2(s)) 2H_(2(s)) O
i.e 2 molecules of hydrogen combine with 1 molecule of oxygen to produce 2 molecules of steam
Example:
1. 60〖cm〗^3of hydrogen are sparked with 20〖cm〗^3 of oxygen at 1000C and 1 atmosphere. What is the volume of the steam produced?
Solution
2H2 + O2 2H2O
From the equation, 2 molecules of hydrogen react with 1 molecule of oxygen to produce 2 molecules of steam.
2H_2 + O2 2H2O
2 vol 1vol 2 vol (combining volumes)
i.e. 2〖cm〗^3 1〖cm〗^3 2〖cm〗^3
From the above information, when 2〖cm〗^3(2 vol) of H_2 react, 1〖cm〗^3(1 vol) of O2will react i.e. half of H_2vol, to give 2〖cm〗^3(2 vol) of H2O.
Thus, 10〖cm〗^3 of H_2 will react with 5 〖cm〗^3 of O2 to produce 10〖cm〗^3 of H2O and so on.
From the question, we have 60〖cm〗^3of H_2 and 20〖cm〗^3 of O2, thus, when all the 20〖cm〗^3 of O2react, only 40〖cm〗^3 ofH_2 will react to give 40〖cm〗^3 of H2O,because the volume of H2 is the same as that of H2O i.e.
2H_2 + O2 2H2O
2 vol 1 vol 2 vol
2〖cm〗^3 1〖cm〗^3 2〖cm〗^3
40〖cm〗^3 20〖cm〗^3 40〖cm〗^3
Thus, the volume of steam (H2O) formed is 40〖cm〗^3
2.What volume of propane is left unreacted when 80〖cm〗^3 of oxygen and 20〖cm〗^3 of propane react according to the equation below?
C3H8(g) + 5O2(g) 3CO2(g) + 4H2O
Solution
C3H8(g) + 5O2(g) 3CO2(g) + 4H2O
1vols 5vols
1〖cm〗^3 5〖cm〗^3
4〖cm〗^3 20〖cm〗^3
Volume of the propane before the reaction =20〖cm〗^3
The volume that reacted =4〖cm〗^3
Volume that did not react= volume before
Reaction – volume that reacted i.e. 20 – 4 =16〖cm〗^3
EVALUATION
State Gay –Lussac’s law
State Avogadro’s law.
50〖cm〗^3of methane were burnt completely in oxygen according to the equation below.
CH4 + 2O2 Co2 + 2H2O
Calculate: (a) volume of oxygen used (b) Volume of carbon(Iv) oxide produced
Volume of steam produced.
GENERAL EVALUATION
OBJECTIVE TEST
A gas occupies 30.0〖dm〗^3 at S.T.P. What volume will itoccupy at 910C and 52662.5〖Nm〗^(-2). (a) 20.0〖dm〗^3 (b) 40.0〖dm〗^3 (c) 60.0〖dm〗^3 (d) 76.96〖dm〗^3 (e) 100.0〖dm〗^3
Gases can be easily compressed because. (a) the molecule are relatively far apart (b) the molecule are quite close together (c) the molecule are very soft (d) the molecules are in constant, rapid motion (e) collusions between gas molecules are elastic.
A give mass of gas occupies X_1 〖cm〗^3 at Y1K. When the temperature is changed to Y2K, the volume becomes X2〖cm〗^3, the pressure remaning constant. Which of the following equations correctly express the relationship between X_1X2Y1andY2?
X_1 (b) X_1Y1 = X2Y2(c)X1/Y1 = X2/Y2 (d) X_1X2 =X_1 Y2(e)X_1 = X2Y1Y2
Kelvin temperature can be converted into Celsius temperature by. (a) 0_C = K – 273 (b) k + 273 (c) (0C+273)/k (d) (k+273)/0C
What will be the new volume (v) if the new pressure is halved and the initial pressure remain the same. (a) 〖2p〗_(1 ) V_1 = p_(2 ) V_2 (b) p_(1 ) V_1 =〖2p〗_(2 ) V_2 (c) (P1 V1)/2 = (P2 V2)/2 (d) p_(1 ) V_1 =(P2 V2)/2
ESSAY QUESTIONS
130〖cm〗^3 of a gas at 200C exert a pressure of 750mmHg. Calculate its volume is increased to 150〖cm〗^3 at 350C
State the laws attributed to Gay- Lussac’s and Avogadro’s respectively. (b) How does Avogadro’s law explain Gay Lussac’s law? Use the formation of steam or Ammonia
Draw the graphical representation of both Boyle’s and Charles’s law respectively.
Convert the following temperature to K. (a) 150C (b)2750C (c) 880C
The volume of gas at 250C (298k) is 100〖cm〗^3. What will be the volume at (a.) 750C (348k) (b). 500C (223)k, pressure remaining constant?
WEEKEND ASSIGNMENT:
Read about Graham’s law, Avogadro’s number and the mole concept.
WEEKEND ACTIVITY:
(a) Define Graham’s law of diffusion.(b) What is mole and mole concept?
From the gas laws, we know that the volume of a gas depends on both its temperature and pressure. The relationship between the three variable; i.e. volume,, temperature and pressure can be summarized up as follows:
If V ∝1/(P ) (Boyle’s law at constant temperature) and V ∝ T (Charle’s law at constant pressure)
V ∝1/(P ) × T (both temperature and pressure may vary) orPV/T = K (a mathematical constant for a fixed mass of gas)
PV/T =K is often known as the general gas equation.
GENERAL GAS EQUATION
General gas equation states that for fixed mass of a gas under any set of conditions of V, P and T, the value of PV/T must remain constant. If for a fixed mass of gas V1 is the volume at pressure
P1 and absolute temperature T1 and V2 is the volume at pressure P2 and absolute temperature T2 it follows that.
P1V1/T =P2V2/T2
The general gas equation can be used to find the volume of a gas when both its pressure and temperature change. Thus;
V2 =( P1 V1 T2)/(P2 V2)
The standard temperature and pressure
The value of gases are sometimes given in standard temperature and pressure (S. T. P). These values are standard temperature= 273k and standard pressure = 760mmHg. The S.I unit of standard pressure when used is 1.01 × 103Nm-2
Examples
At S. T. P a certain mass of gas occupies a volume of 790cm3, find the temperature at which the gas occupies 1000cm3 and has a presence of 720mmHg
P1V1/T1= P2V2/T2
P1 = 760mmHg (at stp), V1= 790cm3
T1 = 273k (at stp), V2 = 1000cm3
P2 = 726mmHg
T2 = New Temperature
∴T2 =(P2 V2 T1)/(P1 V1)
= (720 ×1000 ×273)/(760 ×790) = 330.1k
The new temperature of the gas is 330.1k
A given mass of gas occupies 850cm3at 320k and 0.92 × 103Nm-2
of pressure. Calculate the volume of the gas at S.T.P.
P1V1/T1 = P2V2/T2
P1= 0.92 × 103Nm-2 T1= 320k
V1= 850cm3 P2= SP + 1.01 × 103Nm-2
T2= 273k (at stp)
V2 = new volume of gas.
∴V2 = (P1 V1 T2)/(P2 T1) = (0.92 ×850 ×273)/(1.01 ×〖10〗^3 ×320) = 660.5〖cm〗^3
EVALUATION
Explain the general gas equation.
If the volume of a given mass of a gas at 298k and a pressure of 205.2 ×〖10〗^5 〖Nm〗^(-2) is 2.12〖dm〗^3. what is the volume of the gas S.T.P (standard pressure= 101.3 ×103Nm-2, standard temperature = 273)
SUB TOPIC 3;GAY- LUSSAC’S LAW AND AVOGADRO’S LAW
Gay- Lussac’s law describes the combining volumes of gases that react together.In his experiment, all temperatures and pressures were kept constant:
A. STEAM: Gay- Lussac’s observed that two volumes of hydrogen reacted with one volume of oxygen to yield two volumes of steam
Hydrogen + Oxygen Steam
Volume 2 : 1 2
Ratio 2 : 1 2
B. HYDROGEN CHLORIDE GAS: One volume of hydrogen combined with one volume of chloride to yield two volumes of hydrogen.
Hydrogen + Chlorine Hydrogen Chloride
i.e. H2 + CL2 2HCL
Volume= 1 : 1 2
Ratio = 1 : 1 2
C.Carbon monoxide + Oxygen + CarbonIVoxide
2CO + O2 2CO2
Ratio 2 : 1 : 2
Gay- Lussac’s noticed that the combining volumes as well as the volumes of the products, if gaseous, were related by simple ratios of whole numbers. He proposed the law of combining volume or gaseous volumes.
Hence; Gay- Lussac’s law combining volumes states thatwhen gases react, they do so in volumes which are simple ratios to one another and to the volumes of the producers, if gaseous provide that the temperature and the pressure remain constant.
EXAMPLES
1.What is the volume of oxygen required to burn completely 45〖cm〗^3 of methane (〖CH〗_4)?
Equation of reaction:
〖 CH〗_4(g) + 〖 20〗_(2(g))) 〖Co 〗_(2(g)) + 2〖HO〗_(2(g))
Vol: 1 2 1 2
Ratio:1 2 1 2
By Gay- Lussac’s Law:
1 volume of methane required 2 volumes of oxygen i.e.
1〖cm〗^3 of methane requires 2〖cm〗^3 of oxygen
∴ 45〖cm〗^3 of methane require 90〖cm〗^3 of oxygen
3.20〖cm〗^3 of carbonIIoxide are sparked with 20〖cm〗^3 of oxygen. If all the volumes of gases are measured at a S.T.P, calculate the volume of the residual gases after sparking?
Equation of reaction 2〖Co〗_g+ O_2g 2〖Co〗_2g
Combining volume 2 : 1 : 2
Volumes before sparking 20〖cm〗^3 10〖cm〗^3, 20〖cm〗^3
Volumes after sparking -10 20
Residual gases = un-reacted oxygen + carbon (IV) oxide formed
Volume of residual gas = 10〖cm〗^3+ 20〖cm〗^3= 30〖cm〗^3
AVOGADRO’S LAW
Avogadro’s Law states thatequal volumes of all gases at the same temperature and pressure contain the same number of molecules.
This law means that for any type of gas,e.g. oxygen hydrogen, Chlorine etc if their volumes are the same, they will have the same number of molecules.
Avogadro’s Law is easily applied to convert volume of gases to the number of molecules. Avogadro’s Law can be used to solve problem under Gay –Lussac’s law of combining volumes.
The formation of steam from reaction of Hydrogen and Oxygen is given below:
Reaction Hydrogen + Oxygen Steam
Volume: 2 1 2
Gay –Lussac’s: 2 : 1 : 2
Avogadro’s Law: 2 : 1 : 2
This agrees with the equation below:
2H_(2(g)) + O_(2(s)) 2H_(2(s)) O
i.e 2 molecules of hydrogen combine with 1 molecule of oxygen to produce 2 molecules of steam
Example:
1. 60〖cm〗^3of hydrogen are sparked with 20〖cm〗^3 of oxygen at 1000C and 1 atmosphere. What is the volume of the steam produced?
Solution
2H2 + O2 2H2O
From the equation, 2 molecules of hydrogen react with 1 molecule of oxygen to produce 2 molecules of steam.
2H_2 + O2 2H2O
2 vol 1vol 2 vol (combining volumes)
i.e. 2〖cm〗^3 1〖cm〗^3 2〖cm〗^3
From the above information, when 2〖cm〗^3(2 vol) of H_2 react, 1〖cm〗^3(1 vol) of O2will react i.e. half of H_2vol, to give 2〖cm〗^3(2 vol) of H2O.
Thus, 10〖cm〗^3 of H_2 will react with 5 〖cm〗^3 of O2 to produce 10〖cm〗^3 of H2O and so on.
From the question, we have 60〖cm〗^3of H_2 and 20〖cm〗^3 of O2, thus, when all the 20〖cm〗^3 of O2react, only 40〖cm〗^3 ofH_2 will react to give 40〖cm〗^3 of H2O,because the volume of H2 is the same as that of H2O i.e.
2H_2 + O2 2H2O
2 vol 1 vol 2 vol
2〖cm〗^3 1〖cm〗^3 2〖cm〗^3
40〖cm〗^3 20〖cm〗^3 40〖cm〗^3
Thus, the volume of steam (H2O) formed is 40〖cm〗^3
2.What volume of propane is left unreacted when 80〖cm〗^3 of oxygen and 20〖cm〗^3 of propane react according to the equation below?
C3H8(g) + 5O2(g) 3CO2(g) + 4H2O
Solution
C3H8(g) + 5O2(g) 3CO2(g) + 4H2O
1vols 5vols
1〖cm〗^3 5〖cm〗^3
4〖cm〗^3 20〖cm〗^3
Volume of the propane before the reaction =20〖cm〗^3
The volume that reacted =4〖cm〗^3
Volume that did not react= volume before
Reaction – volume that reacted i.e. 20 – 4 =16〖cm〗^3
EVALUATION
State Gay –Lussac’s law
State Avogadro’s law.
50〖cm〗^3of methane were burnt completely in oxygen according to the equation below.
CH4 + 2O2 Co2 + 2H2O
Calculate: (a) volume of oxygen used (b) Volume of carbon(Iv) oxide produced
Volume of steam produced.
GENERAL EVALUATION
OBJECTIVE TEST
A gas occupies 30.0〖dm〗^3 at S.T.P. What volume will itoccupy at 910C and 52662.5〖Nm〗^(-2). (a) 20.0〖dm〗^3 (b) 40.0〖dm〗^3 (c) 60.0〖dm〗^3 (d) 76.96〖dm〗^3 (e) 100.0〖dm〗^3
Gases can be easily compressed because. (a) the molecule are relatively far apart (b) the molecule are quite close together (c) the molecule are very soft (d) the molecules are in constant, rapid motion (e) collusions between gas molecules are elastic.
A give mass of gas occupies X_1 〖cm〗^3 at Y1K. When the temperature is changed to Y2K, the volume becomes X2〖cm〗^3, the pressure remaning constant. Which of the following equations correctly express the relationship between X_1X2Y1andY2?
X_1 (b) X_1Y1 = X2Y2(c)X1/Y1 = X2/Y2 (d) X_1X2 =X_1 Y2(e)X_1 = X2Y1Y2
Kelvin temperature can be converted into Celsius temperature by. (a) 0_C = K – 273 (b) k + 273 (c) (0C+273)/k (d) (k+273)/0C
What will be the new volume (v) if the new pressure is halved and the initial pressure remain the same. (a) 〖2p〗_(1 ) V_1 = p_(2 ) V_2 (b) p_(1 ) V_1 =〖2p〗_(2 ) V_2 (c) (P1 V1)/2 = (P2 V2)/2 (d) p_(1 ) V_1 =(P2 V2)/2
ESSAY QUESTIONS
130〖cm〗^3 of a gas at 200C exert a pressure of 750mmHg. Calculate its volume is increased to 150〖cm〗^3 at 350C
State the laws attributed to Gay- Lussac’s and Avogadro’s respectively. (b) How does Avogadro’s law explain Gay Lussac’s law? Use the formation of steam or Ammonia
Draw the graphical representation of both Boyle’s and Charles’s law respectively.
Convert the following temperature to K. (a) 150C (b)2750C (c) 880C
The volume of gas at 250C (298k) is 100〖cm〗^3. What will be the volume at (a.) 750C (348k) (b). 500C (223)k, pressure remaining constant?
WEEKEND ASSIGNMENT:
Read about Graham’s law, Avogadro’s number and the mole concept.
WEEKEND ACTIVITY:
(a) Define Graham’s law of diffusion.(b) What is mole and mole concept?
WEEK 3
TOPIC: THE GAS LAWS
CONTENT: (i) Ideal gas equation and Graham’s law
(ii) Molar volume of gases – Avogadro number and the mole concept
(iii)Calculations based on the gas law
SUB-TOPIC 1: IDEAL GAS EQUATION AND GRAHAM’S LAW
The ideal gas: This is a gas sample whose properties correspond, within experimental error, to the relationship PV =RT. An ideal gas must obey all the rules guiding Boyle’s and Charles’s laws. Fourquantities’ are important in all experimental work, measurements or calculations involving gases. They are: (i) volume (ii) pressure (iii) temperature and (iv)numbers of moles or mass.
Three of these appear in the general gas equation:
P1V1/T1= (P2 V2)/T2
The equation states that for an ideal gasPv/T is a constant.
Again, it has also been proven experimentally that equal volumes of all gases at the temperature and pressure contain the same number of molecules, and that one mole of any gas at s.t.p occupies a volume of 22.4〖dm〗^3, This fact together with Boyle’s and Charles’ lead to another equation:
PV=RT
Where R is a constant termed as the molar gas constant.
The above equation is called an ideal gas equation and it is true for all gases. The equation only applies to one mole of gas. For n- mole of gas, we have general form of the equation.
PV = nRT
Where P is in atom, V in〖dm〗^3, T in K, n= no of moles of gases
At s. T. P is in gas
R = Pv/nT = (1(atm) × 22.44 (〖dm〗^3))/(1(mol) × 273(K))
R= 0.082atm〖dm〗^3 k^- 〖mol〗^-
Graham’s law of diffusion
This law states that, the rate at a constant temperature and pressure, the rate of diffusion of a gas is inversely proportional to the square root of its, relative molecular mass i.e.
R1/R2∝√p2/p1Where R1 and R2 are the rates of diffusion and P1 and P2, the densities of the two gases.
The density is directly proportional to its molecular mass.
EVALUATION
State Gram’s law of diffusion in relation to density of a gas.
What is an ideal gas?
Write down the ideal gas equation for n-mole of a gas.
SUB TOPIC 2: Molar volume of gases- Avogadro number and the mole concept.
The molar volume of any gas is the volume occupied by one mole of that gas at s.t.p and is equal numerically to 22.4〖dm〗^3. i.e. one mole of any gas at s.t.p occupies the same volume, the value is 22.4〖dm〗^3. This value is called molecular mass or molar mass.
From the Avogadro’s law, the molar volume for all gases contains the same number of molecules. This number is called the Avogadro’s number or constant and the value is 6.02 ×〖10〗^23 at s.t.p
MOLE of a substance represents 6.02 ×〖10〗^23 particles of any substance. Therefore a mole refers to Avogadro’s number of particles of any substance.
In summary, the molar mass of a gas contains Avogadro’s number of molecules which is 6.02 ×〖10〗^23 and occupies a volume of 22.4〖dm〗^3 at s.t.p.
The atomic mass of every element also contains Avogadro’s number of atoms.
The mole concept- This says that one mole of any substance contains the same number of particles; which can be atoms, molecules or ions. This number is 6.023 ×〖10〗^23 〖dm〗^3 (the Avogadro’s number)
Examples
1.158g of a gas at s.t.p occupies a volume of 5000〖dm〗^3. What is the relative molecular masss of the gas? (Molar volume at s.t.p= 22.4〖dm〗^3 mol-1
Solution
Volume of gas: v= 50.00〖dm〗^3
Molar volume of gas; v= 22.4〖dm〗^3 mol-1
N= amount in moles
=V/V
N= 50/(22.4〖dm〗^(3mol-1) )= 2.23mol
Molar mass M of the gas=
M/n = 158g/(22.4〖dm〗^3 mol-1) = 70.8
Molar mass = 71g/mol
2. What is the mass of 3 moles of oxygen gas O2? (O = 16)
Mass of 1 mole of O2= (2× 16)g =32g
Mass of 3moles of O2= (3 × 32)g = 96g
3. How Many moles are there in 20g of CaCo3? [CaCo3 =100]
Molar mass of CaCo3= 100g
100g of CaCo3= 1 mole
20g of CaCo3 =(20 )/100×1mole = 0.2moles
EVALUATION
1.Using the relationship between mole and Avogadro’s number. Define mole in six ways.
SUB-TOPIC: 3 Calculations based on the Gas law
Boyle’s law
A sample of gas occupied 390〖cm〗^3 at a pressure of 760mm of mercury. Calculate the volume it would occupy at780mm of mercury if the temperature remains constant.
SOLUTION
P1 V1= P2 V2 T = constant
P1 = 760mm
V1 = 390〖cm〗^3
V2= ?
∴ V2= (P1 V1)/P2 = (760mm × )/780mm 390〖cm〗^3
V2= 380〖cm〗^3
What pressure would be required to compress 7.75litres of hydrogen at atmospheric pressure to 5litres?
[1 atm= 760mm, Hg = 76= torrents]
Solution
P1V1= P2 V2 (T constant)
P= l atm
V1= 7.75 litres
P2= ?
V2= 5litres
P1V1= P2 V2
P2=(P1 V1)/V2 = (latin×7.75litres)/(5 lites)
= 1.55atm
CHARLES’ LAW
A certain mass of gas occupies 211〖cm〗^3 at 180C and 740mmHg pressure. What volume will it occupy [still gaseous] at -200C and 770mmHg?
Solution
180C = (18 + 273) k = 291k
-200C = (-20 + 273) k = 253k
Using (P1 V1)/T1 = (P2 V2)/T2
(740mm×211 〖cm〗^3)/291k = (740mm×V2)/253k
∴ V2= (740mm×211〖cm〗^3×253 〖dm〗^3)/(770mm×291k)
V2= 176,3〖cm〗^3
CONTENT: (i) Ideal gas equation and Graham’s law
(ii) Molar volume of gases – Avogadro number and the mole concept
(iii)Calculations based on the gas law
SUB-TOPIC 1: IDEAL GAS EQUATION AND GRAHAM’S LAW
The ideal gas: This is a gas sample whose properties correspond, within experimental error, to the relationship PV =RT. An ideal gas must obey all the rules guiding Boyle’s and Charles’s laws. Fourquantities’ are important in all experimental work, measurements or calculations involving gases. They are: (i) volume (ii) pressure (iii) temperature and (iv)numbers of moles or mass.
Three of these appear in the general gas equation:
P1V1/T1= (P2 V2)/T2
The equation states that for an ideal gasPv/T is a constant.
Again, it has also been proven experimentally that equal volumes of all gases at the temperature and pressure contain the same number of molecules, and that one mole of any gas at s.t.p occupies a volume of 22.4〖dm〗^3, This fact together with Boyle’s and Charles’ lead to another equation:
PV=RT
Where R is a constant termed as the molar gas constant.
The above equation is called an ideal gas equation and it is true for all gases. The equation only applies to one mole of gas. For n- mole of gas, we have general form of the equation.
PV = nRT
Where P is in atom, V in〖dm〗^3, T in K, n= no of moles of gases
At s. T. P is in gas
R = Pv/nT = (1(atm) × 22.44 (〖dm〗^3))/(1(mol) × 273(K))
R= 0.082atm〖dm〗^3 k^- 〖mol〗^-
Graham’s law of diffusion
This law states that, the rate at a constant temperature and pressure, the rate of diffusion of a gas is inversely proportional to the square root of its, relative molecular mass i.e.
R1/R2∝√p2/p1Where R1 and R2 are the rates of diffusion and P1 and P2, the densities of the two gases.
The density is directly proportional to its molecular mass.
EVALUATION
State Gram’s law of diffusion in relation to density of a gas.
What is an ideal gas?
Write down the ideal gas equation for n-mole of a gas.
SUB TOPIC 2: Molar volume of gases- Avogadro number and the mole concept.
The molar volume of any gas is the volume occupied by one mole of that gas at s.t.p and is equal numerically to 22.4〖dm〗^3. i.e. one mole of any gas at s.t.p occupies the same volume, the value is 22.4〖dm〗^3. This value is called molecular mass or molar mass.
From the Avogadro’s law, the molar volume for all gases contains the same number of molecules. This number is called the Avogadro’s number or constant and the value is 6.02 ×〖10〗^23 at s.t.p
MOLE of a substance represents 6.02 ×〖10〗^23 particles of any substance. Therefore a mole refers to Avogadro’s number of particles of any substance.
In summary, the molar mass of a gas contains Avogadro’s number of molecules which is 6.02 ×〖10〗^23 and occupies a volume of 22.4〖dm〗^3 at s.t.p.
The atomic mass of every element also contains Avogadro’s number of atoms.
The mole concept- This says that one mole of any substance contains the same number of particles; which can be atoms, molecules or ions. This number is 6.023 ×〖10〗^23 〖dm〗^3 (the Avogadro’s number)
Examples
1.158g of a gas at s.t.p occupies a volume of 5000〖dm〗^3. What is the relative molecular masss of the gas? (Molar volume at s.t.p= 22.4〖dm〗^3 mol-1
Solution
Volume of gas: v= 50.00〖dm〗^3
Molar volume of gas; v= 22.4〖dm〗^3 mol-1
N= amount in moles
=V/V
N= 50/(22.4〖dm〗^(3mol-1) )= 2.23mol
Molar mass M of the gas=
M/n = 158g/(22.4〖dm〗^3 mol-1) = 70.8
Molar mass = 71g/mol
2. What is the mass of 3 moles of oxygen gas O2? (O = 16)
Mass of 1 mole of O2= (2× 16)g =32g
Mass of 3moles of O2= (3 × 32)g = 96g
3. How Many moles are there in 20g of CaCo3? [CaCo3 =100]
Molar mass of CaCo3= 100g
100g of CaCo3= 1 mole
20g of CaCo3 =(20 )/100×1mole = 0.2moles
EVALUATION
1.Using the relationship between mole and Avogadro’s number. Define mole in six ways.
SUB-TOPIC: 3 Calculations based on the Gas law
Boyle’s law
A sample of gas occupied 390〖cm〗^3 at a pressure of 760mm of mercury. Calculate the volume it would occupy at780mm of mercury if the temperature remains constant.
SOLUTION
P1 V1= P2 V2 T = constant
P1 = 760mm
V1 = 390〖cm〗^3
V2= ?
∴ V2= (P1 V1)/P2 = (760mm × )/780mm 390〖cm〗^3
V2= 380〖cm〗^3
What pressure would be required to compress 7.75litres of hydrogen at atmospheric pressure to 5litres?
[1 atm= 760mm, Hg = 76= torrents]
Solution
P1V1= P2 V2 (T constant)
P= l atm
V1= 7.75 litres
P2= ?
V2= 5litres
P1V1= P2 V2
P2=(P1 V1)/V2 = (latin×7.75litres)/(5 lites)
= 1.55atm
CHARLES’ LAW
A certain mass of gas occupies 211〖cm〗^3 at 180C and 740mmHg pressure. What volume will it occupy [still gaseous] at -200C and 770mmHg?
Solution
180C = (18 + 273) k = 291k
-200C = (-20 + 273) k = 253k
Using (P1 V1)/T1 = (P2 V2)/T2
(740mm×211 〖cm〗^3)/291k = (740mm×V2)/253k
∴ V2= (740mm×211〖cm〗^3×253 〖dm〗^3)/(770mm×291k)
V2= 176,3〖cm〗^3
WEEK 4
GRAHAM’S LAW
100〖cm〗^3of oxygen diffuse through an office in 60 seconds while it takes 120seconds for the same office. Calculate the molecular mass of the unknown gas [0=16]
Solution
R/RxO2 = Mx/MO2
Since the rate of diffusion is inversely proportional to the time taken:
R/RxO2 = =tx/tO2 = √Mx/MO2
(tx/tO2) = Mx/MO2
Mx= MO2 ×(tx/tO2)2= 32 ×(120/60)2= 32 ×2^2
Mx = 32 × 4 = 128g
200〖cm〗^3of hydrogen diffused through a porous pot in 40 seconds. How long will it take 300〖cm〗^3of chlorine to diffuse through the same pot?
Solution
200〖cm〗^3of hydrogen diffused in 40secs
∴ 300〖cm〗^3 of chlorine will diffuse in
(300〖cm〗^3)/(200〖cm〗^3 ) × 4020
(3 × 20) = 60seconds
Now, using the equation
t1/t2 =√M1/M2
Where t= 60s, M1 = molecular mass of hydrogen
i.e H2= (2 ×1) =2
M2= molecular mass of chlorine = cl2 =2× 35.5 = 71
T2 = t1√M1/M2= 60√71/2= 60 √35.5 = 60 × 5.96
= 357.5sec
Time of diffusion of chlorine = 358s.
How many times the rate of diffusion of hydrogen is faster than that of oxygen and what law do you use to get the answer? [ vapour density] of [H=1, O=16]
Solution
Rate (R+) of diffusion of H2=
√((Density of O_2)/(Density of H_2 ))
R1/R2=√(16/1)= R1/R2= 4/1
∴Hydrogen diffuses four times faster. The law used is Graham’s law of diffusion.
RELATIVE VAPOUR DENSITY OF A GASE
The vapour density of a gas or vapour is the number of times a given volume of gas (or vapour) is heavier than the same volume of hydrogen measured and weighed under the same temperature and pressure
Vapour density = (mass of 1 vol of a gas or vapour)/(mass of equal volume of hydrogen)
Applying Avogadro’s law, it is possible to show that the vapour density of a gas is related to the relative molecular mass of the gas.
V.D= (mass of 1 mole of a gas or vapour)/(mass of 1 molecule of hydrogen)
V.D = (mass of 1 vol of a gas )/(mass of 2 atoms of hydrogen)
∴2 x V.D =relative molecular mass
The density of hydrogen at S.T.P is 0.09〖dm〗^3
Example
Calculate the vapour densities of the following gases from the given data.
560〖cm〗^3 of oxygen at S.T.P weighs 0.8g
1,400〖cm〗^3 of sulphur (iv) oxide weighs 4g
Solution
1000m^3 of hydrogen at S.T.P weighs 0.09g
∴ 560〖cm〗^3 of hydrogen at (560〖cm〗^3)/(100〖cm〗^3 )× 0.09
= 0.05g
V.D= massofagivenvolumeofgas/massofeequalvolumeofhydrogen
∴Vapour density of oxygen=
(mass of 560 of oxygen)/(mass of 560 of hydrogen)
1000〖cm〗^3 of hydrogen at S.T.P weighs 0.09g.
∴ 1400 of hydrogen will weighs
(1400 ×0.09)/1000= 0.126g
Vapour density= massofagivenvolumeofgas/massofeequalvolumeofhydrogen
∴Vapour density of SO2= (mass of 1400〖cm〗^3 of SO2)/(mass of 1400〖cm〗^3 of H_2 )
= 4g/0.126= 31.74= 32
EVALUTION
Deduce the relationship between relative molecular mass and vapour density of a substance.
Define vapour density of a gas.
GENERAL EVALUATION
OBJECTIVE TEST
A liquid begins to boil when
Its vapour pressure is equal to the vapour pressure of its solid at the given temperature
Molecules start escaping from its surface
Its vapour pressure equals the atmosphere pressure
It volume is slightly increased.
Hydrogen diffuses through a porous plug
At the same rate as oxygen
Twice as fast as oxygen
Three times as fast as oxygen.
Four times as fast as oxygen
When pollen grains are suspended in water and viewed through a microscope, they appear to be in a state of constant but erratic motion. This is due to:
Convention current (b) small change in temperature (c) a chemical reaction between the pollen grains and the water (d) the bombardment of the pollen grain by molecules of water.
If the quantity of oxygen occupying a 2.76litre container at pressure of 0.825 atmosphere and 300k is reduced by one-half, what is the pressure exerted by the remaining gas?
1.650atm
0.825atm
0.413atm
0.275atm
200〖cm〗^3of oxygen diffused through a porous plug in 50secs. How long will 80〖cm〗^3 of methane (CH4) take to diffuse through the same porous plug under the same conditions (C= 12, O= 16, H=1)
40sec
20sec
14sec
7sec
ESSAY QUESTIONS
1(a) State Graham’s law of diffusion
Arrange the following gases in decreasing order of diffusion rate: Chlorine, hydrogen chloride, hydrogen sulphur and Carbon IV oxide
[ H=1, C= 12, O=16, S= 32, Cl=35.5]
2. What do you understand by s.t.p?
(b) If the volume of a given mass of gas at 298k and pressure of 205.2 ×〖10〗^(3 ) 〖Nm〗^(-2) is 2.12〖dm〗^3, what is the volume at S.T.P? Standard pressure= 101.3 ×〖10〗^(3 ) Nm. Standard temperature= 273k
3. Calculate the number of moles of the following at S.T.P
i. 16g of oxygen
ii. 67.2〖dm〗^3 of nitrogen gas, and
iii. 1.14〖dm〗^3 of hydrogen chloride gas.
O=16, H=14, N=1. Molar volume of gas at S.T.P = 22.4〖dm〗^3
(i) Convert 33℃ and -41℃ to Kelvin scale
(ii) Convert 270k and 315k to 0℃
PRE READING ASIGNMENT:
Read about standard separation techniques.
WEEKEND ACTIVITY:
List all the separation techniques that you know.
100〖cm〗^3of oxygen diffuse through an office in 60 seconds while it takes 120seconds for the same office. Calculate the molecular mass of the unknown gas [0=16]
Solution
R/RxO2 = Mx/MO2
Since the rate of diffusion is inversely proportional to the time taken:
R/RxO2 = =tx/tO2 = √Mx/MO2
(tx/tO2) = Mx/MO2
Mx= MO2 ×(tx/tO2)2= 32 ×(120/60)2= 32 ×2^2
Mx = 32 × 4 = 128g
200〖cm〗^3of hydrogen diffused through a porous pot in 40 seconds. How long will it take 300〖cm〗^3of chlorine to diffuse through the same pot?
Solution
200〖cm〗^3of hydrogen diffused in 40secs
∴ 300〖cm〗^3 of chlorine will diffuse in
(300〖cm〗^3)/(200〖cm〗^3 ) × 4020
(3 × 20) = 60seconds
Now, using the equation
t1/t2 =√M1/M2
Where t= 60s, M1 = molecular mass of hydrogen
i.e H2= (2 ×1) =2
M2= molecular mass of chlorine = cl2 =2× 35.5 = 71
T2 = t1√M1/M2= 60√71/2= 60 √35.5 = 60 × 5.96
= 357.5sec
Time of diffusion of chlorine = 358s.
How many times the rate of diffusion of hydrogen is faster than that of oxygen and what law do you use to get the answer? [ vapour density] of [H=1, O=16]
Solution
Rate (R+) of diffusion of H2=
√((Density of O_2)/(Density of H_2 ))
R1/R2=√(16/1)= R1/R2= 4/1
∴Hydrogen diffuses four times faster. The law used is Graham’s law of diffusion.
RELATIVE VAPOUR DENSITY OF A GASE
The vapour density of a gas or vapour is the number of times a given volume of gas (or vapour) is heavier than the same volume of hydrogen measured and weighed under the same temperature and pressure
Vapour density = (mass of 1 vol of a gas or vapour)/(mass of equal volume of hydrogen)
Applying Avogadro’s law, it is possible to show that the vapour density of a gas is related to the relative molecular mass of the gas.
V.D= (mass of 1 mole of a gas or vapour)/(mass of 1 molecule of hydrogen)
V.D = (mass of 1 vol of a gas )/(mass of 2 atoms of hydrogen)
∴2 x V.D =relative molecular mass
The density of hydrogen at S.T.P is 0.09〖dm〗^3
Example
Calculate the vapour densities of the following gases from the given data.
560〖cm〗^3 of oxygen at S.T.P weighs 0.8g
1,400〖cm〗^3 of sulphur (iv) oxide weighs 4g
Solution
1000m^3 of hydrogen at S.T.P weighs 0.09g
∴ 560〖cm〗^3 of hydrogen at (560〖cm〗^3)/(100〖cm〗^3 )× 0.09
= 0.05g
V.D= massofagivenvolumeofgas/massofeequalvolumeofhydrogen
∴Vapour density of oxygen=
(mass of 560 of oxygen)/(mass of 560 of hydrogen)
1000〖cm〗^3 of hydrogen at S.T.P weighs 0.09g.
∴ 1400 of hydrogen will weighs
(1400 ×0.09)/1000= 0.126g
Vapour density= massofagivenvolumeofgas/massofeequalvolumeofhydrogen
∴Vapour density of SO2= (mass of 1400〖cm〗^3 of SO2)/(mass of 1400〖cm〗^3 of H_2 )
= 4g/0.126= 31.74= 32
EVALUTION
Deduce the relationship between relative molecular mass and vapour density of a substance.
Define vapour density of a gas.
GENERAL EVALUATION
OBJECTIVE TEST
A liquid begins to boil when
Its vapour pressure is equal to the vapour pressure of its solid at the given temperature
Molecules start escaping from its surface
Its vapour pressure equals the atmosphere pressure
It volume is slightly increased.
Hydrogen diffuses through a porous plug
At the same rate as oxygen
Twice as fast as oxygen
Three times as fast as oxygen.
Four times as fast as oxygen
When pollen grains are suspended in water and viewed through a microscope, they appear to be in a state of constant but erratic motion. This is due to:
Convention current (b) small change in temperature (c) a chemical reaction between the pollen grains and the water (d) the bombardment of the pollen grain by molecules of water.
If the quantity of oxygen occupying a 2.76litre container at pressure of 0.825 atmosphere and 300k is reduced by one-half, what is the pressure exerted by the remaining gas?
1.650atm
0.825atm
0.413atm
0.275atm
200〖cm〗^3of oxygen diffused through a porous plug in 50secs. How long will 80〖cm〗^3 of methane (CH4) take to diffuse through the same porous plug under the same conditions (C= 12, O= 16, H=1)
40sec
20sec
14sec
7sec
ESSAY QUESTIONS
1(a) State Graham’s law of diffusion
Arrange the following gases in decreasing order of diffusion rate: Chlorine, hydrogen chloride, hydrogen sulphur and Carbon IV oxide
[ H=1, C= 12, O=16, S= 32, Cl=35.5]
2. What do you understand by s.t.p?
(b) If the volume of a given mass of gas at 298k and pressure of 205.2 ×〖10〗^(3 ) 〖Nm〗^(-2) is 2.12〖dm〗^3, what is the volume at S.T.P? Standard pressure= 101.3 ×〖10〗^(3 ) Nm. Standard temperature= 273k
3. Calculate the number of moles of the following at S.T.P
i. 16g of oxygen
ii. 67.2〖dm〗^3 of nitrogen gas, and
iii. 1.14〖dm〗^3 of hydrogen chloride gas.
O=16, H=14, N=1. Molar volume of gas at S.T.P = 22.4〖dm〗^3
(i) Convert 33℃ and -41℃ to Kelvin scale
(ii) Convert 270k and 315k to 0℃
PRE READING ASIGNMENT:
Read about standard separation techniques.
WEEKEND ACTIVITY:
List all the separation techniques that you know.
WEEK 5
TOPIC: STANDARD SEPERATION TECHNIQUES FOR MIXTURES
SUB-TOPICS:
(a) MIXTURES
(b) SEPERATION OF MIXTURES
(c) TYPES OF SEPERATION TECHNIQUES
(i)Sieving
(ii) Magnetic separation
(iii Sublimation
(iv)Centrifugation
(v) Filtration
MIXTURE: A mixture contains two or more constituents which can easily be separated by physical methods. The constituents of mixtures can be elements, compounds or both.
SEPERATION OF MIXTURES
Most substances are not pure they are mixtures of compounds or elements.To study the properties of a substance a chemist must make sure that the substance is pure.The constituents of a mixture are not chemically combined therefore they can be separated by physical methods.There are various methods by which we can separate pure substances from mixtures.
TYPES OF SEPERATION TECHNIQUES.
1. Separating a mixture of two solids.
(a) SIEVING: This is a method used to separate a mixture of two solids of different sizes.
(b) Magnetic separation: To separate magnetic solids from non-magnetic ones.
(c) Sublimation: Solids which sublime are separated from other solids that sublime (i.e. they turn directly from solid to gas and ffrom gas to solid without turning to liquid) are
Ammonium chloride
Aluminum chloride
Camphor
Iodine crystals
Solid CO2 (dry ice)
Purification by sublimation
2.Separating a mixture of an insoluble solid and a liquid.
(a) DECANTATION: The mixture is allowed to stand for some time until it separates into two distinct layer and an upper clear liquid layer . The clear liquid can be carefully poured or decanted into a second container.
(b) CENTRIFUGATION: This involves the use of a centrifuge to separate solid particles from a liquid as shown in the diagram below. As the centrifuge spins the mixture the solids separates and settles at the bottom of the test tube while the liquids on the top layer can easily be decanted.
(c) Centrifugation process
(d) FILTRATION:As shown below, the mixture is poured into s porous material (filter paper) folded inside a funnel. The solid particle thatremains inside the liquid that drips through the filter paper is known as the filtrate.
Filtration process
EVALUATION:
(a) Explain briefly, how a mixture of sand and water can be separated.
GENERAL EVALUATION
OBJECTIVE TEST
1. A mixture of gari and stones can be separated by (a) filtration (b) centrifugation (c) sieving (d) sublimation
2. A mixture of iodine crystals and common salt can be separated by. (a) Sublimation (b) filtration (c) sieving (d) centrifugation
3. Sieving is a technique used to separate mixtures containing solid particles of (a) small sizes (b) large sizes (c) the same sizes (d) different sizes
4. Which of the following methods can be used to separate a mixture of iron fillings and sulphur? (a) Filtration (b) magnetization (c) sublimation (d) centrifuging
5. The following are subliming substance except. (a)Ammonium chloride (b) Aluminum (c) Sodium chloride (d) Camphor
ESSAY QUESTIONS
(1) Draw a clearly labeled diagram to illustrate separation of a mixture of chalk suspension.
(2) Explain how a centrifuge machine works.
(3) Explain using diagram how you would separate a mixture of sand and ammonium chloride.
(4) List all the methods that can be used to separate an insoluble solid from a liquid.
(5) Fill in the gaps.
A porous material like _________ can be used to separate ______ particles from ___________. After separation the liquid is called ________ while the particles are called ____________.
WEEKEND ASSIGNMENT:
Read about the industrial applications of sieving, magnetic separation, centrifugation, and filtration and write themout.
WEEKEND ACTIVITY:
Mix sand and water together in a container. Allow it to stand for some minutes. What method would you use to separation the water and the sand.
PRE-READING ASSIGNMEN:
Read about distillation and fractional distillation from page 15 and 17 of Osei Yaw Ababio
SUB-TOPICS:
(a) MIXTURES
(b) SEPERATION OF MIXTURES
(c) TYPES OF SEPERATION TECHNIQUES
(i)Sieving
(ii) Magnetic separation
(iii Sublimation
(iv)Centrifugation
(v) Filtration
MIXTURE: A mixture contains two or more constituents which can easily be separated by physical methods. The constituents of mixtures can be elements, compounds or both.
SEPERATION OF MIXTURES
Most substances are not pure they are mixtures of compounds or elements.To study the properties of a substance a chemist must make sure that the substance is pure.The constituents of a mixture are not chemically combined therefore they can be separated by physical methods.There are various methods by which we can separate pure substances from mixtures.
TYPES OF SEPERATION TECHNIQUES.
1. Separating a mixture of two solids.
(a) SIEVING: This is a method used to separate a mixture of two solids of different sizes.
(b) Magnetic separation: To separate magnetic solids from non-magnetic ones.
(c) Sublimation: Solids which sublime are separated from other solids that sublime (i.e. they turn directly from solid to gas and ffrom gas to solid without turning to liquid) are
Ammonium chloride
Aluminum chloride
Camphor
Iodine crystals
Solid CO2 (dry ice)
Purification by sublimation
2.Separating a mixture of an insoluble solid and a liquid.
(a) DECANTATION: The mixture is allowed to stand for some time until it separates into two distinct layer and an upper clear liquid layer . The clear liquid can be carefully poured or decanted into a second container.
(b) CENTRIFUGATION: This involves the use of a centrifuge to separate solid particles from a liquid as shown in the diagram below. As the centrifuge spins the mixture the solids separates and settles at the bottom of the test tube while the liquids on the top layer can easily be decanted.
(c) Centrifugation process
(d) FILTRATION:As shown below, the mixture is poured into s porous material (filter paper) folded inside a funnel. The solid particle thatremains inside the liquid that drips through the filter paper is known as the filtrate.
Filtration process
EVALUATION:
(a) Explain briefly, how a mixture of sand and water can be separated.
GENERAL EVALUATION
OBJECTIVE TEST
1. A mixture of gari and stones can be separated by (a) filtration (b) centrifugation (c) sieving (d) sublimation
2. A mixture of iodine crystals and common salt can be separated by. (a) Sublimation (b) filtration (c) sieving (d) centrifugation
3. Sieving is a technique used to separate mixtures containing solid particles of (a) small sizes (b) large sizes (c) the same sizes (d) different sizes
4. Which of the following methods can be used to separate a mixture of iron fillings and sulphur? (a) Filtration (b) magnetization (c) sublimation (d) centrifuging
5. The following are subliming substance except. (a)Ammonium chloride (b) Aluminum (c) Sodium chloride (d) Camphor
ESSAY QUESTIONS
(1) Draw a clearly labeled diagram to illustrate separation of a mixture of chalk suspension.
(2) Explain how a centrifuge machine works.
(3) Explain using diagram how you would separate a mixture of sand and ammonium chloride.
(4) List all the methods that can be used to separate an insoluble solid from a liquid.
(5) Fill in the gaps.
A porous material like _________ can be used to separate ______ particles from ___________. After separation the liquid is called ________ while the particles are called ____________.
WEEKEND ASSIGNMENT:
Read about the industrial applications of sieving, magnetic separation, centrifugation, and filtration and write themout.
WEEKEND ACTIVITY:
Mix sand and water together in a container. Allow it to stand for some minutes. What method would you use to separation the water and the sand.
PRE-READING ASSIGNMEN:
Read about distillation and fractional distillation from page 15 and 17 of Osei Yaw Ababio
WEEK 6
WEEK 5
TOPIC: STANDARD SEPARATION TECHNIQUES FOR MIXTURE
CONTENT:
(a) Separating a soluble solid from a liquid by
Evaporation
Crystallization
Fractional crystallization
Precipitation
(b) Separating a liquid from a solution.
Distillation
(c) Separating a mixture of two or more liquids.
Fractional distillation
Separating funnel
SEPARATING A SOLUBLE SOLID FROM A LIQUID:
Evaporation: Evaporation method is used to recover a solid solute from a solution. The solvent(liquid) is usually sacrificed
Evaporation process
Note: Evaporation method is not suitable for salts that can easily be destroyed by heating.
(a) CRYSTALLIZATION: Crystallization is a method used to separate salts which decompose easily on heating from their solutions.The salt solution (the mixture) is heated to drive away some of the liquid (i.e. to evaporate some of the liquid) (i.e. to evaporate some of the liquid) until solution becomes concentrated or saturated. The concentrated solution remaining is allowed to cool slowly resulting in the formation of crystals. Crystal formation can be induced by (i) adding crystals of the same salt to serve as seed. (ii) Scratching the inside of the vessel containing the solution.
Note: If all the liquid is evaporated a powder will be obtained and not crystals. This powder might also contain impurities which otherwise would have remained in the solution and not contaminate the crystals. Many crystals formed on cooling saturated solution contain water which is chemically combined and loosely bonded to the crystals.This water is called water of crystallization. Salts which contain water of crystallization are said to be hydrated.Those which do not are anhydrous. Those are often powders.
EVALUATION:
Write four examples of salts that contain water of crystallization.
(b) FRACTIONAL CRYSTALLIZION:
This is a method used to separate a mixture containing different soluble solid solutes in a liquid. The solubility of the different solid solutes in the given solvent must differ at different temperatures. The process of separation is the same as in crystallization process. While cooling the solution crystals of the relevant solid solutes will come out of the solution leaving behind the others which are still within their limits of solubility.
PRECIPITATION: This method is used to separate a solid which has a difference in solubility in two different miscible liquids. For example, ethanol and water are two miscible liquids. Iron(II)tetraoxosulphate(vi) is soluble in water but not in ethanol.On addition of ethanol to a solution containing a mixture of iron(ii)tetraoxosulphate and water, the iron(ii)tetraoxosulphate(vi) will be precipitated out and can be separated by filtration.
Precipitation can also be said to be a process whereby dissolved solid solutes can be removed from a solution using a precipitating agent e.g. the technique of precipitation can be employed to remove Hardness in water as well as to purify water. Here the precipitating agent e.g. Lime (calcium hydroxide, Ca(OH)2) is added into a sample of the hard water (mixture) and the solid solute can then settle down at the bottom of the container as precipitating. The solid particle (precipitate) can then be removed by filtration.
In the purification of an impure water using Alum. If a piece of an impure water using Alum. If a piece of Alum (Potassium – aluminumtetraoxosulphate (vi), KAl (SO4)2. 12 H2O) is added to an impure water the impurities immediately are precipitated at the bottom of the container. Clear water will be left on top. This water can be called by filtration or decantation.
(1) SEPARATING A LIQUID FROM A SOLUTION (Impure liquid)
DISTILLATION
Distillation is the evaporation of water or other liquids from a solution and its recovery on a pure state by condensation.The method of distillation is used to recover a solvent (liquid from a solution (mixture). That is, a pure liquid from an impure liquid(mixture). The apparatus used are shown in the diagram below.
Simple distillation process
At the end of the distillation process the liquid that is collected at the end of the Liebig condenser is called the distillate. The solutes and other impurities are left behind in the distillation flask.
Difference between evaporation and distillation
DISTILLATION EVAPORATION
Mainly obtaining the solvent Mainly for obtaining salt from solution.
It involves boiling and condensation It involves boiling only.
EVALUATION:
Explain briefly the process of distillation.
(2) SEPARATING A MIXTURE OF TWO OR MORE LIQUIDS
(a) To separate miscible liquids
FRACTIONAL DISTILLATION
Fractional Distillation is a process used to separate or mixture of miscible liquids by a repeated evaporation and condensation making use of fractionating column (as shown in the diagram below)
Mixture of two or more miscible liquids are separated into, its component parts. The liquids distil according to their boiling points starting with the liquid with the lowest boiling point. The apparatus used is the same as in distillation except for the presence of a fractionating column between the flask and the condenser.
Note: For efficient fractional distillation, the difference in the boiling points between successive fractions must be more than 100c.
EVALUATION:
1.Explain briefly, the process of fractional distillation.
(b) Separating immiscible liquids (using separating funnel method)
This a method used to separate a mixture of immiscible liquids e.g. a mixture of petrol and water. When the two liquids are added together they do not mix, instead they separate into two distinct layers, a lower denser layer and an upper less dense layer in the funnel as below.
EVALUATION:
Draw a labeled diagram to show how you would separate a mixture of kerosene and water.
GENERAL EVALUATION
OBJECTIVE TEST:
(1) Separating funnel is used to separate one of the following mixtures. (a) Ethanol and water (b) Iodine and salt (c) petrol and water (d) sand and water
(2) Fractional distillation of petroleum depends on differences in (a) Molar mass (b) densities (c) freezing points (d) boiling point
(3) Fractional distillation is used to separate (a) an insoluble substance from a soluble volatile substance (b)substances which are absorbed differently and which differ in their solubility in a solvent(c) Liquids with differing boiling points (d) Gas, Liquid or solid impurities from a mixture
(4) A mixture of sand, ammonium chloride and sodium chloride is best separated by. (a) Sublimation followed by addition of water and filtration (b) Sublimation followed by addition of water and evaporation (c) addition of water followed by crystallization and sublimation.
(5) Separating funnel is used for separating a mixture of (a) Liquids with different boiling points (b) sediments of liquid. (c) Liquids with different colours. (d) liquids that are immiscible
ESSAY QUESTIONS:
(1) Name the most suitable physical method for each of the following. (a) Containing groundnut oil from a mixture of the oil and water. (b) Obtaining pure water from sea water.
(2) Draw the laboratory set up most suitable for each of the following. (a) Separating of a mixture of palm oil and water (b) Separate of pure liquid from an impure liquid.
(3) State one industrial application of each of the following methods of separation explaining clearly the procedure. (a) crystallization (b) filtration (c) fractional distillation (d) evaporation
(4) With the aid of a labelled diagram only show how pure sample of ethanol (alcohol) can be obtained from a mixture of ethanol and water.
(5) Why is sodium chloride solution regarded as a mixture? (b) Draw a labelled diagram to show how pure sodium chloride can be obtained from its solution.
WEEKEND ASSIGNMENT:
Read about recrystallization from page 5 of comprehensive certificate chemistry; write out the procedures.
WEEKEND ACTIVITY:
Get some impure water in a container and try to purify it by using Alum. How will you separate the pure liquid after the precipitating process?
PRE-READING ASSIGNMENT.
Read about different types of chromatographic methods from page 18 of New Certificate Chemistry by Osei Yaw Ababio.
TOPIC: STANDARD SEPARATION TECHNIQUES FOR MIXTURE
CONTENT:
(a) Separating a soluble solid from a liquid by
Evaporation
Crystallization
Fractional crystallization
Precipitation
(b) Separating a liquid from a solution.
Distillation
(c) Separating a mixture of two or more liquids.
Fractional distillation
Separating funnel
SEPARATING A SOLUBLE SOLID FROM A LIQUID:
Evaporation: Evaporation method is used to recover a solid solute from a solution. The solvent(liquid) is usually sacrificed
Evaporation process
Note: Evaporation method is not suitable for salts that can easily be destroyed by heating.
(a) CRYSTALLIZATION: Crystallization is a method used to separate salts which decompose easily on heating from their solutions.The salt solution (the mixture) is heated to drive away some of the liquid (i.e. to evaporate some of the liquid) (i.e. to evaporate some of the liquid) until solution becomes concentrated or saturated. The concentrated solution remaining is allowed to cool slowly resulting in the formation of crystals. Crystal formation can be induced by (i) adding crystals of the same salt to serve as seed. (ii) Scratching the inside of the vessel containing the solution.
Note: If all the liquid is evaporated a powder will be obtained and not crystals. This powder might also contain impurities which otherwise would have remained in the solution and not contaminate the crystals. Many crystals formed on cooling saturated solution contain water which is chemically combined and loosely bonded to the crystals.This water is called water of crystallization. Salts which contain water of crystallization are said to be hydrated.Those which do not are anhydrous. Those are often powders.
EVALUATION:
Write four examples of salts that contain water of crystallization.
(b) FRACTIONAL CRYSTALLIZION:
This is a method used to separate a mixture containing different soluble solid solutes in a liquid. The solubility of the different solid solutes in the given solvent must differ at different temperatures. The process of separation is the same as in crystallization process. While cooling the solution crystals of the relevant solid solutes will come out of the solution leaving behind the others which are still within their limits of solubility.
PRECIPITATION: This method is used to separate a solid which has a difference in solubility in two different miscible liquids. For example, ethanol and water are two miscible liquids. Iron(II)tetraoxosulphate(vi) is soluble in water but not in ethanol.On addition of ethanol to a solution containing a mixture of iron(ii)tetraoxosulphate and water, the iron(ii)tetraoxosulphate(vi) will be precipitated out and can be separated by filtration.
Precipitation can also be said to be a process whereby dissolved solid solutes can be removed from a solution using a precipitating agent e.g. the technique of precipitation can be employed to remove Hardness in water as well as to purify water. Here the precipitating agent e.g. Lime (calcium hydroxide, Ca(OH)2) is added into a sample of the hard water (mixture) and the solid solute can then settle down at the bottom of the container as precipitating. The solid particle (precipitate) can then be removed by filtration.
In the purification of an impure water using Alum. If a piece of an impure water using Alum. If a piece of Alum (Potassium – aluminumtetraoxosulphate (vi), KAl (SO4)2. 12 H2O) is added to an impure water the impurities immediately are precipitated at the bottom of the container. Clear water will be left on top. This water can be called by filtration or decantation.
(1) SEPARATING A LIQUID FROM A SOLUTION (Impure liquid)
DISTILLATION
Distillation is the evaporation of water or other liquids from a solution and its recovery on a pure state by condensation.The method of distillation is used to recover a solvent (liquid from a solution (mixture). That is, a pure liquid from an impure liquid(mixture). The apparatus used are shown in the diagram below.
Simple distillation process
At the end of the distillation process the liquid that is collected at the end of the Liebig condenser is called the distillate. The solutes and other impurities are left behind in the distillation flask.
Difference between evaporation and distillation
DISTILLATION EVAPORATION
Mainly obtaining the solvent Mainly for obtaining salt from solution.
It involves boiling and condensation It involves boiling only.
EVALUATION:
Explain briefly the process of distillation.
(2) SEPARATING A MIXTURE OF TWO OR MORE LIQUIDS
(a) To separate miscible liquids
FRACTIONAL DISTILLATION
Fractional Distillation is a process used to separate or mixture of miscible liquids by a repeated evaporation and condensation making use of fractionating column (as shown in the diagram below)
Mixture of two or more miscible liquids are separated into, its component parts. The liquids distil according to their boiling points starting with the liquid with the lowest boiling point. The apparatus used is the same as in distillation except for the presence of a fractionating column between the flask and the condenser.
Note: For efficient fractional distillation, the difference in the boiling points between successive fractions must be more than 100c.
EVALUATION:
1.Explain briefly, the process of fractional distillation.
(b) Separating immiscible liquids (using separating funnel method)
This a method used to separate a mixture of immiscible liquids e.g. a mixture of petrol and water. When the two liquids are added together they do not mix, instead they separate into two distinct layers, a lower denser layer and an upper less dense layer in the funnel as below.
EVALUATION:
Draw a labeled diagram to show how you would separate a mixture of kerosene and water.
GENERAL EVALUATION
OBJECTIVE TEST:
(1) Separating funnel is used to separate one of the following mixtures. (a) Ethanol and water (b) Iodine and salt (c) petrol and water (d) sand and water
(2) Fractional distillation of petroleum depends on differences in (a) Molar mass (b) densities (c) freezing points (d) boiling point
(3) Fractional distillation is used to separate (a) an insoluble substance from a soluble volatile substance (b)substances which are absorbed differently and which differ in their solubility in a solvent(c) Liquids with differing boiling points (d) Gas, Liquid or solid impurities from a mixture
(4) A mixture of sand, ammonium chloride and sodium chloride is best separated by. (a) Sublimation followed by addition of water and filtration (b) Sublimation followed by addition of water and evaporation (c) addition of water followed by crystallization and sublimation.
(5) Separating funnel is used for separating a mixture of (a) Liquids with different boiling points (b) sediments of liquid. (c) Liquids with different colours. (d) liquids that are immiscible
ESSAY QUESTIONS:
(1) Name the most suitable physical method for each of the following. (a) Containing groundnut oil from a mixture of the oil and water. (b) Obtaining pure water from sea water.
(2) Draw the laboratory set up most suitable for each of the following. (a) Separating of a mixture of palm oil and water (b) Separate of pure liquid from an impure liquid.
(3) State one industrial application of each of the following methods of separation explaining clearly the procedure. (a) crystallization (b) filtration (c) fractional distillation (d) evaporation
(4) With the aid of a labelled diagram only show how pure sample of ethanol (alcohol) can be obtained from a mixture of ethanol and water.
(5) Why is sodium chloride solution regarded as a mixture? (b) Draw a labelled diagram to show how pure sodium chloride can be obtained from its solution.
WEEKEND ASSIGNMENT:
Read about recrystallization from page 5 of comprehensive certificate chemistry; write out the procedures.
WEEKEND ACTIVITY:
Get some impure water in a container and try to purify it by using Alum. How will you separate the pure liquid after the precipitating process?
PRE-READING ASSIGNMENT.
Read about different types of chromatographic methods from page 18 of New Certificate Chemistry by Osei Yaw Ababio.
WEEK 7
TOPIC: STANDARD SEPARATION TECHNIQUES FOR MIXTURES.
SUB-TOPICS:
Separating complex mixtures by chromatography
Floating
fro station(froth flotation)
Pure and impure substances
Test for purity
Separating complex mixtures by chromatography: This is a method of separation of the components of mixtures of solutes from a solution (mixture) using a solvent (liquid) moving over a porous, adsorbent medium e.g. filter paper or gel. This method can be mixtures f soluble substances.There are different types of chromatographic methods. Paper chromatography (ascending paper chromatography), column chromatography, thin layer chromatography and gas chromatography.
ASCENDING PAPER CHROMATOGRAPHY
As shown in the above diagram, the apparatus include: a glass jar with lid, filter paper, clips, solvent (water or ethanol). The solution containing the mixture of solutes to be separated is spotted unto the strips of paper near one end.
The paper is then suspended in a closed air- tight jar with the spotted end (but not the spot) dipping into the solvent. As the solvent ascends the paper the different solutes in the mixture gets dissolved and also more along the paper strip at different speeds and hence become separated. The paper strip is removed from the jar when the solvent has moved about three-quarters way up the strip. It is dried and if necessary sprayed with appropriate chemical reagents to locate the positions of the various along the strip. Each solute can then be identified by the distance it has traveled.This is done by comparing its distance with those of known standard substances.
EVALUATION
1.Draw a labelled diagram to show how the component of a black dye can be separated.
FLOATATION:
Floatation method is based on the wide difference in the densities of the components of the mixture. The method is used for the separation of a mixture of two solids in which one component is light and the other is heavy. On the addition of a liquid in which neither is soluble, one component sinks, while the other floats. e.g. a mixture of coarse sand and wooden cork.
PROCEDURE: Place the mixture in a beaker and add plenty of water. The sand particles sink, while the wooden corks float.
FROTH FLOATATION (Frostation)
This method is specifically used to separate an ore of a metal from earthly impurities.
PROCEDURE: The ore is crushed into powder and then mixed with water containing detergent, in order to cause frothing (foaming).
Air is then blown into the mixture so that the earthly impurities sink while the ore floats and mixes with the foam.The ore is finally recovered from the foam
PURE AND IMPURE SUBTANCES: The following are the criteria for purity of chemical substances.
DENSITY: The density of a pure substance is definite and constant, while that of an impure substance higher than expected.
MELTING POINT: The melting point of a pure solid is sharp and definite. The presence of an impurity lowers the melting point of a substance, and spread its melting point over a wide range of temperature.
FREEZING POINT: The freezing point of a pure liquid is sharp and definite; the presence of an impurity lowers the freezing point.
BOILING POINT: The boiling point of a pure liquid is sharp and definite. An impurity raises the boiling point of a pure liquid.
TEST FOR PURITY
After separation of substances from mixtures, it is important to know if they are pure. A pure solid should melt at a constant temperature. A pure liquid should boil at a constant temperature. A pure dye should give only one spot on a chromatogram. The melting points or boiling points of pure substances are fixed. These temperatures change if impurities are present. To assess the purity of a substance its melting point (if it is a solid) or its melting point ( if it is liquid) is determined (if the value obtained agrees with that given in a book of data, then the substance is pure).
The apparatus below can be used to find the melting point of a solid.
The melting point of a solid is the temperature at which it changes to liquid.The melting point tube is very thin- a capillary tube- and the substance under test must be finely powdered so that it can be packed into the capillary tube (melting point tube). The beaker containing the oil is heated slowly and the oil stirred vigorously. If the solid is pure it will all melt at a constant temperature. i.e. it will have a sharp melting point.
NOTE: If impurities are present the mixture will melt slowly over a range of temperatures below the melting point of the pure solid.
EVALUATION:
1.How will you know that a given liquid is not pure?
DETERMINATION OF THE BOILING POINT OF LIQUIDS
(a) Flammable liquids (b) In flammable liquids
The boiling point of a liquid is the temperature at which its vapour pressure equals atmospheric pressure.
The apparatus shown above can be used to find the boiling points of liquids.
A pure sample of liquid will boil at a fixed temperature and the reading on the thermometer will remain constant. If the light is not pure it will boil over a range of temperature above the boiling point of the pure liquid.
Impurities lower the melting point of a substance and raise its boiling point.
EVALUATION:
1.List five pieces of apparatus that are common to the determination of melting and boiling points of a chemical substance.
SUB-TOPICS:
Separating complex mixtures by chromatography
Floating
fro station(froth flotation)
Pure and impure substances
Test for purity
Separating complex mixtures by chromatography: This is a method of separation of the components of mixtures of solutes from a solution (mixture) using a solvent (liquid) moving over a porous, adsorbent medium e.g. filter paper or gel. This method can be mixtures f soluble substances.There are different types of chromatographic methods. Paper chromatography (ascending paper chromatography), column chromatography, thin layer chromatography and gas chromatography.
ASCENDING PAPER CHROMATOGRAPHY
As shown in the above diagram, the apparatus include: a glass jar with lid, filter paper, clips, solvent (water or ethanol). The solution containing the mixture of solutes to be separated is spotted unto the strips of paper near one end.
The paper is then suspended in a closed air- tight jar with the spotted end (but not the spot) dipping into the solvent. As the solvent ascends the paper the different solutes in the mixture gets dissolved and also more along the paper strip at different speeds and hence become separated. The paper strip is removed from the jar when the solvent has moved about three-quarters way up the strip. It is dried and if necessary sprayed with appropriate chemical reagents to locate the positions of the various along the strip. Each solute can then be identified by the distance it has traveled.This is done by comparing its distance with those of known standard substances.
EVALUATION
1.Draw a labelled diagram to show how the component of a black dye can be separated.
FLOATATION:
Floatation method is based on the wide difference in the densities of the components of the mixture. The method is used for the separation of a mixture of two solids in which one component is light and the other is heavy. On the addition of a liquid in which neither is soluble, one component sinks, while the other floats. e.g. a mixture of coarse sand and wooden cork.
PROCEDURE: Place the mixture in a beaker and add plenty of water. The sand particles sink, while the wooden corks float.
FROTH FLOATATION (Frostation)
This method is specifically used to separate an ore of a metal from earthly impurities.
PROCEDURE: The ore is crushed into powder and then mixed with water containing detergent, in order to cause frothing (foaming).
Air is then blown into the mixture so that the earthly impurities sink while the ore floats and mixes with the foam.The ore is finally recovered from the foam
PURE AND IMPURE SUBTANCES: The following are the criteria for purity of chemical substances.
DENSITY: The density of a pure substance is definite and constant, while that of an impure substance higher than expected.
MELTING POINT: The melting point of a pure solid is sharp and definite. The presence of an impurity lowers the melting point of a substance, and spread its melting point over a wide range of temperature.
FREEZING POINT: The freezing point of a pure liquid is sharp and definite; the presence of an impurity lowers the freezing point.
BOILING POINT: The boiling point of a pure liquid is sharp and definite. An impurity raises the boiling point of a pure liquid.
TEST FOR PURITY
After separation of substances from mixtures, it is important to know if they are pure. A pure solid should melt at a constant temperature. A pure liquid should boil at a constant temperature. A pure dye should give only one spot on a chromatogram. The melting points or boiling points of pure substances are fixed. These temperatures change if impurities are present. To assess the purity of a substance its melting point (if it is a solid) or its melting point ( if it is liquid) is determined (if the value obtained agrees with that given in a book of data, then the substance is pure).
The apparatus below can be used to find the melting point of a solid.
The melting point of a solid is the temperature at which it changes to liquid.The melting point tube is very thin- a capillary tube- and the substance under test must be finely powdered so that it can be packed into the capillary tube (melting point tube). The beaker containing the oil is heated slowly and the oil stirred vigorously. If the solid is pure it will all melt at a constant temperature. i.e. it will have a sharp melting point.
NOTE: If impurities are present the mixture will melt slowly over a range of temperatures below the melting point of the pure solid.
EVALUATION:
1.How will you know that a given liquid is not pure?
DETERMINATION OF THE BOILING POINT OF LIQUIDS
(a) Flammable liquids (b) In flammable liquids
The boiling point of a liquid is the temperature at which its vapour pressure equals atmospheric pressure.
The apparatus shown above can be used to find the boiling points of liquids.
A pure sample of liquid will boil at a fixed temperature and the reading on the thermometer will remain constant. If the light is not pure it will boil over a range of temperature above the boiling point of the pure liquid.
Impurities lower the melting point of a substance and raise its boiling point.
EVALUATION:
1.List five pieces of apparatus that are common to the determination of melting and boiling points of a chemical substance.
WEEK 8
GENERAL EVALUATION
OBJECTIVE TEST
1. The chromatographic separation of ink is based on the ability of the component to.
A: react with each other
B: react with the solvent
C: dissolve in each other in the column
D: move at different speeds in the column
2. The criteria to verify the purity of a solid substance are. I boiling point II melting point III density IV refractive index
A: I, II B: II, III C: I, II, III D: I, II, III, IV
3. A pure dye will A: have a constant boiling point B: have many spots on a chromatogram C:separate from camphor by evaporation method
4. A flammable liquid A: can be heated directly with flame when it is in a container.
5. The best method to separate a mixture of black ink is
A: floatation B: frostation C: ascending paper chromatography D: evaporation
ESSAY QUESTIONS
1. (a) Define the term’ chromatography’. (b) Name the different types of chromatography (c) Describe with the aid of a diagram how you would separate a mixture of inks.
2. (a) List three physical properties that are common criteria for purity of substances. (b) List five pieces of apparatus that are common to the determination of melting and boiling points of a chemical substances
3. (a) State the importance of the measurement of melting and boiling points. (b) Explain briefly why salt is always sprinkled can the icy roads in countries where the temperature falls below 273k.
4. Explain the following term briefly (i) floatation (ii) frostation
5. Draw a labelled diagram only to illustrate the determination of the boiling point of a flammable liquid.
WEEKEND ASSIGNMENT:
Read column chromatography on page 9 of comprehensive certificate chemistry by Ojokuku et al. Explain briefly (with diagram) how chlorophyll in grass extract can be separated by column chromatography.
WEEK ACTIVITY:
Get a beaker, a pair of scissors, water, filter paper and a black ink. Carry out simple ascending paper chromatography according to the set up.
DIAGRAM
Only the ‘tail’ will be dipped inside the water. Observe what happens.
OBJECTIVE TEST
1. The chromatographic separation of ink is based on the ability of the component to.
A: react with each other
B: react with the solvent
C: dissolve in each other in the column
D: move at different speeds in the column
2. The criteria to verify the purity of a solid substance are. I boiling point II melting point III density IV refractive index
A: I, II B: II, III C: I, II, III D: I, II, III, IV
3. A pure dye will A: have a constant boiling point B: have many spots on a chromatogram C:separate from camphor by evaporation method
4. A flammable liquid A: can be heated directly with flame when it is in a container.
5. The best method to separate a mixture of black ink is
A: floatation B: frostation C: ascending paper chromatography D: evaporation
ESSAY QUESTIONS
1. (a) Define the term’ chromatography’. (b) Name the different types of chromatography (c) Describe with the aid of a diagram how you would separate a mixture of inks.
2. (a) List three physical properties that are common criteria for purity of substances. (b) List five pieces of apparatus that are common to the determination of melting and boiling points of a chemical substances
3. (a) State the importance of the measurement of melting and boiling points. (b) Explain briefly why salt is always sprinkled can the icy roads in countries where the temperature falls below 273k.
4. Explain the following term briefly (i) floatation (ii) frostation
5. Draw a labelled diagram only to illustrate the determination of the boiling point of a flammable liquid.
WEEKEND ASSIGNMENT:
Read column chromatography on page 9 of comprehensive certificate chemistry by Ojokuku et al. Explain briefly (with diagram) how chlorophyll in grass extract can be separated by column chromatography.
WEEK ACTIVITY:
Get a beaker, a pair of scissors, water, filter paper and a black ink. Carry out simple ascending paper chromatography according to the set up.
DIAGRAM
Only the ‘tail’ will be dipped inside the water. Observe what happens.