FORM III TOPIC 9. COMPOUNDS OF METALS:


CHEMISTRY FORM THREE QUESTIONS AND ANSWERS 

1. Identify the letter of the correct answer.

     (i) Which of the following is not a compound of metal?

    A: magnesium oxide  B: aluminium oxide

    C: sulphur dioxide   B: iron (ii) iron (ii) sulphate

    (ii) They used to soften water and manufacture glass.

         A: carbonate compounds  B: Nitrate compounds

  (iii) Prepared by direct combination of metal and water.

        A: oxides   B: hydroxides   C: Nitrates  D: sulphates.

  (iv) Carbon dioxide, water and salts are formed when they react with acid.

      A: sulphates   B: chlorides C: nitrates  D: Carbonates

 (v) Action of metals and dilute hydrochloric acid lead to formation of

      A: chlorine   B: chlorides  C: sulphide   D: sulphate

 (vi) Which of the following compounds give oxides when heat?

      A: copper nitrate and silver nitrate   B: zinc nitrate and potassium carbonate

      C: calcium nitrate and copper nitrate  D: lead nitrate and silver nitrate

  (vii) One of the following compounds can be used to remove hardness of water

         A: calcium chloride   B: Aluminium sulphate   C: Potassium hydroxide  D: aluminium nitrate

   (viii) Does not give a metal oxide on heating.

          A: copper nitrate   B: lead nitrate  C: calcium nitrate   D: silver nitrate.

  (ix)  Aluminium sulphate is used to  _____ impurity particles that found in water.

       A: filter   B: dissolve  C: evaporate   D: coagulate.

   (x)  _____ and _____ sulphates are insoluble in water.

       A: Barium, lead  B: sodium, potassium  C: lead, aluminium   D: magnesium, aluminium

 2. Match items of list A  and  B

   (i) Direct method of preparing metal hydroxides.

   (ii) Method of preparing soluble metal carbonates.

   (iii) Indirect method of preparing metal oxides.

   (iv) Compounds of metals.

   (v) Indirect method of preparing metal hydroxides.

   (vi) Preparation of metal chlorides.

   (vii) Method of preparing soluble metal sulphates.

   (ix)  Direct method of preparing metal oxides.

   (x) Insoluble metal carbonates preparation.

  A: Double decomposition method.

  B: Action of metal carbonate, hydroxide or oxides with dilute sulphuric acid.

  C: Direct heating of metals in air.

  D: Have metals and nitrate group.

  E: Action of metals and dilute hydrochloric acid.

  F: A precipitation reaction in a salt solution or solution of alkali.

  G: Passing excess carbon dioxide gas through a solution of a carbonate.

  H: Thermal decomposition of carbonates, hydroxides or nitrates.

   I:  Calcium carbide is used.

  J: Direct combination of metal and water.

  K: Formed when metals undergo chemical reactions with other substances.

  L: solution of soluble salt and that of alkali are used.

 3. List down at least three (3) chemical properties of the following metal compounds.

  4. Analyse at least three (3) uses of the following metal compounds.

 5. With examples, explain the meaning of amphoteric oxides.

3. (a) Chemical properties of metal carbonates

      -Some are soluble and some are insoluble in water.

     - They react with acids to form salts.

      Na2CO3  +  2HCl ----> 2NaCl + H2O + CO2

    - They decompose on heating to form metal  oxide and carbondioxide exept potassium carbonate and sodium carbonate.

   CaCO3 --------> CaO + CO2.

  (b) some of the chemical properties of oxides include the following.

         -they react with water to form metal hydroxides.

        -they react with dilute acids to form salts and water

        -some metal oxides (amphoteric oxides) react with alkalis to form salts

4. (a) some of the uses of metal nitrates are

         -used as fertilizer(NaNO3).

         -used to make photographic film

      (b) some of the uses of metal sulphates include

         - used as plaster cast(CuSO4).

         - used to manufacture ink(FeSO4)

         - used to make pigments and medicines

 5. Amphoteric oxides are metal oxides that may react with both acids and alkalis(bases).

     Examples of amphoteric oxides are

      They react with dilute acids to form normal salts.

              ZnO  +  H2SO4 ------> ZnSO4 + H2O

     They react with alkali to form complex salts.

               ZnO  +  2NaOH ------> Na2ZnO2 + H2O

FORM III TOPIC 8. EXTRACTION OF METALS:

                   
CHEMISTRY FORM THREE QUESTIONS AND ANSWERS 
QUESTIONS.
 1. Identify the most correct answer.
     i) The series of elements arranged in order of increase of electrode potential.
   A: periodic table  B: reactivity series
   C: groups.             D: period
ii) Ionize by electron loss.
   A: positive charge  B: metals  C: halogens      D: non-metals
(iii) Metals that found as oxides
       A: are of medium reactivity
       B: are very very reactive
       C: are least reactive
        D: lead and gold
 (iv) Obtaining metal from its ore
        A: it is mixture separation
        B: it is extraction of metal
        C: Concentration of metal ore
        D: purification of metal
 (v) It occurs as free element
       A: aluminium  B: lead  C: sodium  D: platinum
 (vi) Extracted from water soluble salt
        A: sodium chloride  potassium chloride
        C: sodium.           D: silver
 (vii) The Down's electrolytic cell is the apparatus for extracting:-
         A: sodium B: aluminium  C: platinum  D: iron
 (viii) Loss of electrons, oxygen or increase of oxidation number
          A: oxidation  B: properties of metals
          C: properties of non-metals  D:reduction
  (ix) Bauxite and cryolite are metal ores for
         A: aluminium  B: sodium  C: calcium  D: potassium
  (x) Bayer process is the process of extracting
        A: copper  B: aluminium  C: sulphur  D: potassium
2. Match list  A  and  list  B
          List  A
  i) Evaporating metals by heating leaving behind impurities.
 ii) Impurities escapes as oxides when impure metal exposed to the air.
 iii) Pure metal ore obtained.
 iv) Results to deposition of pure metal on the cathode after impure metal being dissolved in an electrolyte.
 v) Ore is heated in presence of air below its melting point.
 vi) Ore is heated in absence of air below its melting point.
 vii) Chemically, the metal ore is not decomposed when impurities are removed from it.
 viii) Crude metal.
  ix) Haematite
        List  B
  A: When metal ore concentrated
  B: Dressing of metal ore
  C: Impure metal
  D: Refining metal by electrolysis
  E: Metal destruction
  F: Modification of metals
  G: Distillation of metal
  H: Boiling of metals
  I:  Ore that iron extracted from.
  J: Ore that magnesium extracted
  K: Calcination
  L: Roasting
  M: Refining by electrolysis
  N: Refining by oxidation
3.  Explain at least three physical and four chemical properties of metals.
4. Sodium, magnesium, zinc, copper and silver are arranged in order of reactivity series, in which sodium being the most reactive metal.
(a) Which one of these metals is:-
     (i) likely to tarnish most rapidly when it is exposed to the air?
   (ii) most likely to be found in nature?
   (iii) least likely to react with steam?
  (b) Name two metals that extracted by electrolysis of their molten chlorides and state reason for this method.
5. What is the meaning of the following terms and statements.
 
     i) Metal ores
    ii) Extraction of metals
   iii) Coke
6. With examples name and explain types of methods used in extraction of metals.
7. The following is a diagram which shows extraction of sodium


   (a) Name the diagram.
   (b) Name the ore used in extraction of sodium
   (c) label parts of diagrams indicated by letters.
    (d) Name the compound used to lower the melting point of the ore mentioned in 7(b) above.
    (e) Explain, how sodium extracted by using apparatus named in 7(a) above.
8.(a) Explain the Bayer process extracting  aluminium.
   (b) Mention the uses of aluminium.
9. (a) List down the stages used in extraction of copper.
    (b) Mention importance of copper.
10. (a) Observe the diagram below which shows extraction of iron then answer the questions.


  
(a) Name the ore which used
(b)Which substances are fed into furnace at point A?
(c) Name the substance introduced at point C.
(d) Write the balanced chemical reaction taking place at
      i) 1600°C
     ii) 700°C
    iii) 250°C
(e) What is the function of Carbon monoxide formed in the blast furnace?
(f) Name the products formed at D and  E.
(g) Why CaCO3 is added in a blast furnace?
                       ANSWERS.
     i) B  ii)  B  iii) A.  iv) B  v) D  vi) C  vii) A   viii) A  ix) A  x) B
    i) G  ii) N  iii) A  iv)  D  v) L  vi)  K  vii)  B  viii)  C  ix) I
3. - Physical properties of metals.
    a) They are malleable
             Because they can be hammered into thin sheets

FORM III TOPIC 7. CHEMICAL KINETICS, EQUILIBRIUM AND ENERGETICS


CHEMISTRY FORM THREE QUESTIONS AND ANSWERS 
FORM III TOPIC 7. CHEMICAL KINETICS, EQUILIBRIUM AND ENERGETICS:

            QUESTIONS:
1. Identify the most correct answer at each item from the alternatives provided.
i) Describes rate and factors that cause chemical reaction and affect rate of chemical reaction.
   A: equilibrium reaction
   B: reversible reaction.
   C: chemical kinetics
   D: equilibrium position
 ii) Lowering of activation energy.
      A: decrease rate of chemical reaction
      B:  make reacting particles to start reaction easily.
      C: favours backward reaction
      D: stops the chemical reactions.
 iii)  According to the Collision theory, chemical reaction only takes place when particles
    A: collide  B: break up  C: bond. Warmed
 iv) Even if reacting particles contact, they will not react if
    A: they do not posses activation energy
    B: they are cold
    C: they are few in number
    D: temperature increases
 v) It favours forward reaction when temperature lowered
    A: reaction of hot chemicals
    B: acid base reaction
    C: endothermic reaction
    D: exothermic reaction
vi) It proceed in both forward and back ward directions.
    A: chemical reactions  B: chemical equilibrium
    C: irreversible reaction  D: reversible reaction
 vii) Change of either temperature, pressure or concentration to the chemical reaction that is at equilibrium may cause
     A: shift of equilibrium position
     B: increase of time for completing the reaction
     C: increase of rate of reaction
     D: high consumption of reactants
 viii) A simple reversible reaction
        A: forward reaction is fast than backward direction
        B: both forward and backward directions proceed at the
             Same rate
         C: change direction with change of conditions
         D: its backward direction is fast than forward direction
  ix) Increase of temperature to the chemical reaction
   
      A: makes reacting particles closer
      B: makes reacting particles disorganised
      C: increase kinetic energy to the reacting particles
      D: increase potential energy to the reacting particles
  x) At a very law temperature the reaction rate slows down even if the reaction is
      A: exothermic  B: endothermic  C: at high temperature  D: reversible
2. Match list A and list B
             List A
       i) The contact process
      ii) The Haber process
     iii) Dissolving concentrated sulphuric acid in water
     iv) Has effect to equilibrium reactions that involves gases
      v) Speed  up rate of chemical reaction
     vi) Irreversible reaction
    vii) Equilibrium point
    viii) increases of temperature
      ix) Rate of chemical reaction
       x) Reversible reaction and equilibrium reaction
                List B
     A: proceed in only one direction
     B: proceed in either direction
     C: take place in the closed systems
     D: water formation
     E: obtained when the rate of forward dire
          ction is equal to that of backward dire
          ction
    F:  favours forward direction in endothermic reaction.
    G: increase production of ammonia gas.
    H: 1/time
    I:  sulphur trioxide production
   J: a  catalyst
   K:  exothermic reaction
   L: pressure.
   M: ammonia gas production
3. Explain the meaning of the following words:-
       a) Chemical catalyst
       b) Reversible reaction
       c) Rate of chemical reaction
       d) Irreversible reaction
       e) Endothermic reaction
       f)  Exothermic reaction
    
4. State Le  Chatelier's principle.
5. a) Mention factors that affect rate of chemical reaction.
     b) Explain how factors mentioned in(5 a) above affect rate of chemical reaction.
6. Distinguish endothermic reaction from exothermic reaction.
7. List down the factors that affect equilibrium position.
8. What will happen to the equilibrium position?
       


Change of energy=-Ve.                                                    
 (a) if temperature is lowered to the system 
    (b) if some ammonia gas removed from the system
(c) if pressure increased
9. Draw energy level profiles for endothermic reaction and exothermic reactions.
10. Form III students performed an experiment to determine rate of reaction from volume of hydrogen gas produced against time when zinc granules reacted with dilute hydrochloric acid and the following results were obtained.

(a)  Draw graph of volume against rate       (b) Calculate the rate of the chemical reaction.
(c) At what time high volume of hydrogen gas collected?
d) At what time did the experiment produced the largest volume of oxygen?
11. State one example for endothermic reaction and one for exothermic reaction.
12. Observe the chemical equation the state the effect of increase of pressure to the equilibrium position since the system is at equilibrium.

13.
             Explain the effects of the conditions to the production of sulphur trioxide gas from reaction of sulphur dioxide and oxygen if the system is at equilibrium.
      (a) increase of temperature
      (b) increase of pressure
       (c) removing some sulphur dioxide.
       (d) adding a catalyst to the system.
14. State conditions that enable us to know if the chemical reaction is taking place.
        ANSWERS:
1.
   i) C       vi) D
  ii) B.     vii) A
 iii) A     viii) B
 iv) A       ix) C
 v)  D        x) A
2. i) I          vi)A
   ii) M      vii) E
  iii) K      viii) F
  iv) L        ix) H
   v) J         x) C
3. a) is a substance which alters the rate of the chemical reaction and remain unchanged at the end of the reaction.
    b) is a chemical reaction that proceed both forward and backward direction depending varying conditions such as temperature,pressure and concentration.
   c) refers to ratio of the amount of substance used or produced to the time taken.
   d) is a reaction that proceed in one direction.
   e) is a reaction that absorbs energy from the surrounding in form of heat.
  f) is reaction which releases energy to the surrounding in form of heat.
4. It states that "if the system is at equilibrium and one of the factors is changed, the equilibrium position shifts to minimise the effect of change and new equilibrium is established".
5. (a) factors which affect rate of chemical reaction are:-
          -temperature
          -concentration
          -surface area
          -catalyst
     (b) Temperature
                 It affects rate of chemical reaction by increasing kinetic enerqy to reacting particles when it is increased to the system.
            If the chemical reaction is endothermic, its rate increase with increase of temperature(vice versa is true)
            If  the chemical reaction is exothermic, its rate increase with decrease of temperature(vice versa is true)
           Concentration.
              It affects the rate of chemical reaction by increasing or lowering reacting particles.
            The rate of the chemical reaction increase with increase of concentration( number of reacting particles). Vice versa is true.
               Surface ares.
                 It affects rate of chemical reaction depending on the size of chemical substances.
          Small size of chemical substance(powdered) makes higher surface area than the relative large size(granules)
          The rate of chemical reaction increases with increase of surface area(vice versa is true)
              Catalyst.
                 It affects the rate of chemical reaction by speeding up the reaction rate.
      Presence of catalyst in the chemical reaction increase rate of chemical reaction.(vice versa is true)
6. -Endothermic reaction absorbs heat energy from  the surrounding WHILE  Exothermic reaction release heat energy to the surrounding.
    -In an endothermic reaction, the products are formed at relative hing temperature WHILE in an exothermic reaction, the products are formed at relative low temperature.
     -Endothermic reaction indicated by positive symbol of its energy change. (change of energy= +Ve)  WHILE Exothermic reaction indicated by negative symbol of its energy change( change of energy= -Ve)
7. Factors that affect equilibrium position are:-
       -Temperature
       -Concentration
       -pressure
     TEMPERATURE
   Effect of temperature if the reaction is endothermic.
                   When temperature increased to the system it favours forward reaction and if the           Was at equilibrium, increase of temperature shifts equilibrium position to the right
           In order to minimise the effect of change in the system.
   Effect of temperature if the reaction is endothermic.
            Lowering of temperature favours forward reaction hence equilibrium position shifts
            or moves to right to minimise the effect of change in the system.
  CONCENTRATION.
         Concentration affects equilibrium position by favouring the side which have high concentration.
         If there is high concentration to the reactant side forward reaction is favoured and equilibrium position shifts to the right. Vice versa is true.
   PRESSURE.
        Effects of pressure to the system depends on total volume or number of moles at each side of the reaction.
         If at the product side(right) consists fewer number of moles , forward reaction is favoured and equilibrium position shifts to the right. Vice versa is true.
8. (a) lowering of temperature increase rate of forward reaction since the chemical reaction is exothermic.
       Equilibrium position shifts forward.
   (b) if some of ammonia gas removed from the system, the rate of backward reaction because at the left side concentration is lowered.
       Equilibrium position shifts backward.
   (c) increase of pressure speed up rate of forward reaction.
         Equilibrium position shifts backward because it is a side that produce fewer molecules.
     NB. Pressure has no effect if both sides of the equilibrium position have equal number of molecules.
9

10. (a)
     (b) rate is calculated from the graph by obtaining the gradient as shown in 10 (a) above.
              Gradient=rate
              If the slope is steep it means the rate of reaction is high.
    (c) at 180 second, high volume of the gas collected.
    (d) in the first 20 seconds  the largest volume of the gas produced because its slope is steep .
  NB.
        The scales of the graph are easily obtained by taking the highest value of the data dividing by the number of the lines that you prefer to use.
          For example the highest value on vertical line is 66.
           But i prefer to use 18 lines of the graph
                       To get scale interval.
                                   66/18=4
          4cm3 is the interval between two lines(one square)
11. - examples of endothermic Reactions are
                 *photosynthesis process
                 *water vapour formation process
       - examples of exothermic reactions are
                  *explosion
                   * reaction of water an concentrated acid.
                   * combustion of fuel.
12. There is no effect to the equilibrium position because both sides of the equilibrium system have equal volume(number of moles).
13. (a) increase of temperature favours backward reaction since the reaction is exothermic. This makes low production of sulphur trioxide.
       (b) increase of pressure favours forward reaction because to the product side there is fewer moles and low volume. This makes high production of sulphur trioxide gas.
       (c) removal of some sulphur dioxide cause backward reaction due to decrease of concentration to the reactants side. This makes low production of sulphur trioxide gas.
       (d) addition if a catalyst to the system does not affect the production of the sulphur trioxide gas because catalyst will speed up both forward reaction and backward reaction.
14. The conditions that enable us to know if the chemical reaction is taking place are.
        -colour change of the mixture
        -smell
        -precipitation
        -temperature change

FORM III TOPIC 6. IONIC THEORY AND ELECTROLYSIS

CHEMISTRY FORM THREE QUESTIONS AND ANSWERS 
FORM III TOPIC 6. IONIC THEORY AND ELECTROLYSIS

1. Identify the letter correct answer from the alternatives provided.
i) One of the following is not an electrolyte:-
A) A very concentrated sulphuric acid
B) Molten aluminium oxide
C) Copper sulphate solution. 
D) Sodium chloride solution.

ii) Ionic compounds are electrolyte but when they are in solid state they can not conduct electricity because ____
A) Thy do not contain charged particles.
B) They contain charged particles.
C) Ions are not free to move 
D) There is free movement of ions.

iii) Any substance which found in molten state and can allow electricity to pass through is called
A) Electrolyte
B) Conductor
C) Insulator       
D) Electric wire.

iv) Positively and negatively charged particles are called
A) Electrodes
B) Cations
C) Cathodes 
D) Ions

v) _____is elecrolytic unit of charge.
A) Current
B) Faraday 
C) Coulumb
D) Ampere

vi) During electrolysis _____ and ___ ions are discharged when  dilute sodium chloride solution used as an electrolyte and platinum electrodes.
A) Cl and Na
B) O and OH   
C) H and OH
D) H and Cl

vii) An electric current was passed through concentrated solution of hydrochloric acid using carbon electrodes. The substance liberated at anode was.
A) Copper
B) Sodium
C) Hydrogen 
D) Chlorine

viii) Amount of charge carried by one mole of electrons.
A) Faraday
B) 96500
C) Current 
D) Ions

ix) Relates the quantity of electricity passed and amount of substance liberated or deposited at the electrodes.
A) Laws of electrolysis
B) Faraday's constant
C) Electrochemical series
D) Relative molar mass

x) Migrates to positively charged electrode.
A) Copper
B) Sodium
C) Cation
D) Anion.

2. Match items in list A and items in list B.

LIST A.
i) Electrodes
ii) Conductor
iii) Electrolyte
iv) Insulator
v) Anode
vi) Cathode
vii) Weak electrolyte
viii) Strong electrolytes
ix)  Cation
x) Anion

LIST B.
A: Moves in molten state.
B: Slightly ionised in aqueous solution.
C: Do not allow electricity to pass through.
D: Allows electricity to pass through when it is in molten state.
E: Electricity enter or leave electrolytes through them.
F: Only current pass through them
G: Mineral acids and salts
H: Positively charged element.
I: Negatively charged element.
J: Solid substance that allow electricity to pass through.
K: Negatively charged electrode.
L:  Positively charged electrode.
M: Ethanol and water.
N: Migration of ions.

3. Write true for correct statement and false for incorrect statement.
i) Ethanol is a weak electrolyte.

ii) Non-electrolytes exist in form of molecules and do not ionize.

iii) When structure of an electrolyte destroyed ions become free to move.

iv) During electrolysis cations attracted to anode because they are negatively charged.

v) Discharge means losing property of being charged.

vi) During electrolysis chemical reactions produced at electrodes

vii) Concentration of electrolyte is among of factors for the ion to be discharged during electrolysis.

viii) Sodium chloride is a strong electrolyte.

ix) Mass liberated by one Coulomb during electrolysis refers to molar mass.

x) Electrolysis applied in crop production

4. Define the following terms:-

     e) Cathode

5. Explain how ions migrate during electrolysis when an electricity is passing through copper two sulphate solution.

6. State the meaning of the following words in electrolysis.
    a) Weak electrolyte
    b) Conductor
    c) Cation
    d) Strong electrolyte
    e) Anion.

   g) Electrochemical equivalent.


7. Explain reasons for
    a) Sugar.
   b) A very concentrated sulphuric acid.
 to be incapable to conduct electricity while water can conduct.

8. Calculate number of electrons of Aluminium used to deposit 1mole of of Al during electrolysis when molten Aluminium oxide used as electrolyte and 3 Faradays were needed to deposit one mole of  Aluminium .

9. State Faraday's laws of electrolysis.


10. A steady current of 2A passed through a solution containing ions of divalent metal M (M2+)for nine minutes. The mass of metal M liberated was 0.3552g. calculate relative atomic mass of metal M.

11. Write chemical equations for  discharging process at cathode and anode when dilute sulphuric acid is electrolysed using platinum electrodes.

12. A bluish copper(ii) sulphate aqueous solution was electrolysed by copper electrodes.

     (a) Write ionic equations for the reactions.

     (b) Explain what will happen for blue colour of copper(ii) sulphate as electrolysis continue when carbon electrodes used.

13. A solution of sodium hydroxide was electrolysed using platinum electrodes and the solution became alkaline.

(i) write reactions that took place at electrodes.

(ii) explain why the solution remains alkaline.

14.A current of 0.5A were made to flow through silver voltameter for 30 minutes.

Calculate the mass of silver deposited and its equivalent weight.

15.  298500 coulumbs were required to deposit one mole of metalic element Q from its aqueous salt solution. Calculate the valence of Q.

16.  Current of 5A  passed through molten aluminium chloride for 3.5 hours, what number of coulumbs will be?

17. What mass of copper will be liberated during electrolysis of copper sulphate solution by charge of one faraday.

18. A sample of impure silver with a mass of 3.45g was used as anode during electrolysis purifying process. The cathode of pure gold with mass 6.45g used. After electrolysis the cathode found to have 9.66g.

(i) calculate number of coulumbs passed

(ii) what is percentage purity of the impure silver?

19. An element X has a relative atomic mass of 88, when a current of 0.5A was passed through the fused chloride of X for 32 minutes and 10 seconds 0.44g was deposited at cathode.

   (i) Calculate the number of faradays needed to liberate 1 mole of X.

   (ii) Write formula for X ion.

   (iii) write formula for chloride ion.

20. Name ions that move to:-

21. A solution of 1M copper (ii) chloride was electrolysed using graphite electrodes.

   Write the reactions took place at

22. Differentiate chemical equivalent from electrochemical equivalent.

23. A current of 10000 A is passing through an electrolytic cell for 12 hours in order to purify copper.

   (a) What mass of pure copper is deposited at cathode?

    (b) Calculate mass of aluminium that would be deposited if the cell was changed to refine aluminium.

24. 0.02 moles of electrons were passed through a solution of sodium hydroxide using platinum electrodes.
(i) state gases produced
(ii) Calculate number of mole of each gas and volume of each gas at S.T.P

25. A light bulb uses a current of 0.6A  how many Faraday will be used by light by each hour?

26. State the meaning of the following terms in electrolysis.

27. What are the differences between anode and cathode.

28. Electric current was passed through a solution of sodium hydroxide using platinum electrodes, draw a labelled electrolytic cell for this electrolysis. indicate the directions of movement of ions.

SOLUTIONS.

1. i)  A   ii)  C  iii)  A  iv)  D  v)  C
   vi)  C  vii)  D  viii)  B  ix)  A  x) D

2. i) E  ii) J iii) D  iv)  C  v)  L
  vi)  K  vii)  B viii)  G  ix) H  x) I

3. i) False             vi)True
   ii) True              vii)True
  iii) True              viii)True
  iv) False              ix)False
   v) True                x) False

4. a) Electrodes:-  are metal rods or plates through which electric current  enter or leave the electrolytes.

     b)Electrolyte:- is a substance that when it is in solution form or molten state conducts electricity.

     c) Electrolysis:-  defined as decomposition of the compound which is in solution or molten state by passing electricity through it.

    d) Anode:- is a positively electrode which leads electrons out of the electrolyte.

   e) Cathode:- is a negatively electrode which leads electrons into the electrolyte.

5. Copper (ii) solution as electrolyte it contain the following ions:-

    Sulphate ions(So4 2+) and

During electrolysis, Sulphate ions(So4 2+) and Hydroxide ions(-OH) Move to positively electrode or anode.

Copper ions(Cu 2+) and Hydrogen ions(+H) move to negatively electrode or cathode.

6. a) Weak electrolyte:- is an electrolytic compound that ionizes partially in aqueous solution.

   b) Conductor:- is any solid substance which allows electricity.

   b) Cation:- is a positively charged ion.

   c) Strong electrolyte:- is an electrolytic compound that ionizes completely in aqueous solution.

   d) Anion:- is a negatively charged ion.

   e) Electrochemical equivalent:- is the mass of one element liberated by one coulomb of electricity during electricity.

  f) Non- electrolyte :- is the compound which does not conduct electricity in solution or molten state.

  g) Electroplating:-  refers to the coating of metal with a layer of another metal by means of electrolysis.

7.  a) An electrolytic compound conducts electricity due to presence of free ions when it is in solution or molten state but sugar solution contains only molecules not ions that is why it does not conduct electricity.

   b) A very concentrated mineral acids contain only molecules of acids not ions. When become diluted, their molecular structure destroyed and ions formed.

There fore a very concentrated sulphuric acid contains molecules not ions, this is reason for it to be not able to conduct electricity.

    Electrolytic reaction indicates that Aluminium ions discharged and deposited at cathode.

3 moles(Faradays) of electrons needed 1 mole of Aluminium atom to be deposited.

  3 electrons used.

Number of electrons = number of moles of aluminium x Avogadro's constant.

     LA•=Avogadro's constant=6.02x10^23

 The number of electrons of Aluminium=18.06 x10^23

9. There are two faraday's laws of electrolysis which are:-

   a) Faraday's first law of electrolysis and

    b) Faraday's second law of electrolysis

a) Faraday's first law of electrolysis states that " The mass of substance producedc or dissolved at the electrode during electrolysis is proportional to the moles of quantity of electricity transferred at the electrode".

b) Faraday's second law of electrolysis states that " when the same quantity of electricity is passed through the solutions of different electrolytes, the mass of the substances liberated or deposited at the electrodes is directly proportional to their chemical equivalents".

      Data given:  mass liberated=0.3552g

            Mass=ZIt   Z=equivalent weight

                   Z=R.A.M/charge(coulumbs)

      Mass=(R.A.M/ charge) It.

Then.     R.A.M=Mass x charge/It.

       Since 2Faradays required 1 mole of metal M to be liberated.

    And 1mole (faraday)=96500 coulumb(charge)

    96500 is faraday's constant.

       Charge = number of Faradays x Faraday's constant

     Faraday's constant = 96500 coulumbs

   R.A.M=0.3552 x 2 x 96500/2 x 540

Relative atomic mass of metal M =63.5

11. Dilute sulphuric acid as electrolyte contains the following ions:-

             H+ from water and sulphuric acid

           So4 2- from sulphuric acid

During electrolysis H+ migrate to cathode where by OH- and So4 2- migrate to anode.

Hydrogen ions(H+) discharged to form hydrogen gas

        2H+(aq)  +   2e-  ------> H2(g).

Hydroxide ions(OH-) discharged in preference to sulphate ions(So4 2-) because hydroxide ions have greater tendency of losing electrons than sulphate ions. Oxygen formed and collected as gas

     4OH-(aq)  -------> 2H2O(l)  + O2(g).

12.      (a)  Reaction at cathode

Copper ions discharged in preference to hydrogen ions and deposited as brown coat at electrode.

    Cu2+(aq)  +  2e-  --------> Cu(s).

Neither copper ions nor hydroxide ions discharged but copper anode dissolves to form copper ions.

   Cu(s) --------> Cu2+(aq)  +  2e-

       The mass of cathode increases while that of anode decreases.

   (b) Fading of blue colour of copper ions takes place due to decrease of copper ions, finally the colourless solution of dilute sulphuric acid obtained after all copper ions being discharged.

13. i) The ions present in dilute sodium hydroxide are:-

        OH- from water and Na OH.

Hydrogen ions discharged in preference to sodium ions. Hydrogen gas formed

    2H+(aq)   +  2e-  -------> H2(g)

Hydroxide ions discharged. Water and oxygen formed.

     4OH(aq), -----------> 2H2O(aq)  +  O2(g).

  ii)  The solution remains alkaline because sodium ions reacts with hydroxide ions from water molecules formed during reaction which takes place at anode.

14.               Solution.

           Given data
              Current = 0.5 A
            Time=30 min.=1800 seconds
             Mass = required

From
           Mass=zIt
 Where by z= electrochemical equivalent

                    I= current
                    t= time in seconds

First we have to find value of z

        From
                    z= R.A.M/ charge
           R.A.M of silver = 108
Reaction at cathode shows that

1Faraday required to deposit 1 mole

    Ag+  +  e-  -----> Ag

Charge = number of Faraday x Faraday's constant

Charge=1F x 96500
             = 96500 coulumbs

           Z=108/96500
             =0.00112

   mass=0.00112 x 0.5
           =1.007g
              1.007g of silver deposited.

  Equivalent weight (z) =R.A.M/Charge
where by R.A.T = 108
Charge=No. Faradays x F. Constant
            = 1 x 96500
            = 96500 coulumbs
          Z=108/96500
            =0.001119~ 1.119 x 10^-3.

15.       solution
       Given data
       Charge=298500 coulumbs

1 Faraday(mole/electron)=96500 coulumbs

      1F  -------> 96500 coulumbs
       ? <-------- coulumbs.="" nbsp="" p="">
    ? =298500/96500
       =3.093 ~ 3
There fore:-
     3Faradays were required to deposit 1 mole of metalic element Q.

 NB:
  -If 1F required to deposit 1 mole of metalic element( that metal is monovalent)

   -If 2F used to deposit 1 mole of metalic element(that is divalent element)

Element Q is trivalent because its 3 Faradays were required to deposit one mole.

Q3+(aq)  +  3e- -----> Q(s)

    Its valence is 3

But time must be changed into seconds

                    3.5 x 60 x 60= 12600 seconds

 Copper ions discharged  and its mass discharged at cathode.

            Cu2+ (aq)  + 2e- ------> Cu(s)

             2Faredays required to deposit 1 mole of copper.

                  Mass of impure silver = 3.45g   Mass of cathode after electrolysis=9.66g

Mass of cathode(gold) before electrolysis = 6.45g

      Number of coulumbs = required.

During electrolysis the following reaction takes place at cathode.

      Ag+(aq)    +     e-  -------> Ag(s).

It means 1 Faraday required to deposit 1 mole of silver.

Atomic mass of silver = 108g

Mass of silver deposited = 9.66g - 6.45g

 No. of coulumbs=No. F x F.constant

(ii)%purity=M.of pure/M.of ipure x 100

      Mass of pure silver = 9.66g - 6.45g

      Mass of impure silver =  3.45g

                 %purity         = 3.21/3.45  x 100


19. (i)               Solution.

                  Given data
          R.A.M = 88
          Mass  = 0.44g
         Time   = 32 min + 10 seconds = 1930 sec.
From
         Mass = zIt
 
         Z.       = mass/It = 0.44/ 0.5x 1930
                   = 0.000456

        Z.       = R.A.M/coulumbs(No. Of Faradays x Faraday's constant)
        Z.       = R.A.M/ No. Of F x Faraday's constant

No. Faradays = R.A.M/Z x Faraday's constant.

                        = 88/0.000456 x 96500
                        = 2
   2 Faradays were required to liberate 1 mole of of X.

  (ii) since 2 Faradays were required to liberate one mole of X
 
           Number of Faraday can represent number of electrons or moles.
   
           X2+   +    2e- --------> X
     
         The formula for X ion is  X2+.

(iii) Chlorine ionizes by gaining one electron
        Its ionic formula is Cl-

21. Since the solution is dilute, preferential discharge determined by position in electrochemical series.

Hydrogen ions discharged in preference to copper ions

   2H+(aq)  +   2e-  --------> H2(g)

Hydroxide ions discharged in preference to chlorine ions.

           4OH-(aq) -------> 2H2O(l) + O2(g)

22.  Electrochemical equivalent is the mas of of an element liberated by one coulumb of electricity during electrolysis.

                  Chemical equivalent is the mass of an element deposited or liberated by one Faraday during electrolysis.

                Time = 12hr = 12 x 60 x 60 sec.

   Z= R.A.M/No. of Faradays x F.constant

 2F required to deposit 1mole of copper

      Mass= 0.000329 x 10000 x 43200
               =142128g(answer)

         Mass of aluminium= required

   Where by E = chemical equivalent

  Let E1 be chemical equivalent of copper

  E2 be chemical equivalent of aluminium

           E = R.A.M/ No. Of faradays.

            E1= 63.5/2
                = 31.75

            E2= 27/3
                = 9

      31.75/9 = 142128/ mass of Al

      31.75 mass of Al=9 x 142128

       mass of Al = 1279152/31.75
                       = 40288.252g(answer)

24. (i) Ions discharge determined by position of element in electrochemical series because sodium hydroxide solution is dilute.

Hydrogen ions discharged in preference to sodium ions.

         2H+(aq)  +  2e-  ---------> H2(g)

Hydroxide ions discharged and are only anions which found in the solution.

            4OH-(aq)  -------> 2H2O(l)  +  O2(g)

   Gases produced are hydrogen and oxygen.

  (ii)  volume of hydrogen gas at s.t.p

       Hydrogen produced at cathode.

    2H+(aq)  +  2e- -------> H2(g)

It means 2 moles used to produce 1mole of hydrogen gas.

   1mole = 22.4 litres  according molar volume of gases at s.t.p

Volume of hydrogen gas at s.t.p=0.224L.

Oxygen gas produced at anode

    4OH-(aq) ---------> 2H2O(l) + O2(g).

Volume of oxygen at s.t.p =0.112 L.

25.                  Solution.

         Given data
                Current = 0.6A
           Time= 1hour = 60 minutes

In order to calculate number of faradays, quantity of charge (coulumbs )must be obtained.

Coulumbs(charge) = It
But time must be in seconds =  60 x 60
                              = 3600 seconds

          Coulumbs  =  0.6 x 3600
                              =  2160 plumb's

                                   1 F   =  96500
                                    ?      =  2160
                                            = 0.0224 F

 -electrons leave ele   -electrons enter ele

ctrolyte through it.      ctrolyte.

 -it is positively cha    - it is negatively cha

 -it attracts anions.    - it attracts cations.
                         
29. NaOH electrolyte contains, sodium ions(Na+), hydrogen ions (H+) and hydroxide ions, when electricity allowed to pass through it hydrogen and sodium ions move to cathode where by hydroxide ions move to anode.

FORM III TOPIC 5. VOLUMETRIC ANALYSIS.

CHEMISTRY FORM THREE QUESTIONS AND ANSWERS 
FORM III TOPIC 5. VOLUMETRIC ANALYSIS.

QUESTIONS.
1. Identify the letter of the most correct answer.
   i)In a standard solution the following are known:-
      A: mass and moles  B: molarity and concentration
      C: amount of solute and solvent D: mass and molar mass

  ii)Apparatus for measuring accurate volume of liquids are:-
      A: pipette and burette  B: beaker and dropper
      C: Conical flask and reagent bottle D: measuring cylinder and funnel

 iii)The quantity of substance in volumetric analysis expressed in the following forms except:-
      A: number of moles present in certain volume of solution
      B: mass of solute dissolved to make known volume of solution.
      C: molarity
      D: number of particles

  iv)Both methyl orange and phenolphthalein indicators are not used when titrating
      A: weak base against weak acid  B: strong base against strong acid
      C: strong acid against alkaline solution. D: weak base against strong acid

  v)It involves weighing, preparation of solution and calculation
      A: quantitative analysis.  B: volumetric analysis
      C: chemical equations.    D: chemical kinetics

 vi)When water added to the basic compound, the mixture becomes:-
      A: heterogeneous  B: saturated. C: alkaline  D: unsaturated

 vii)Sources of errors when conducting titration experiments may include
       A: reading errors and timing burette tape for the last drop.
       B: mixing error
       C: noise
       D: impurities

 viii)Adding acid from burette into the conical flask containing alkaline solution with indicator is known as:-
      A: titration  B: standardization. C: emulsion  D: basicity

  ix)Any amount of solute dissolved in a solvent to make one litre of a solution is called
      A: molar solution. B: standard solution. C: stock solution  D: molarity

    x) The last drop of solution during titration that lead to the colour change indicates:-
       A: End point of the reaction. B: starting point of the reaction
       C: the reaction is not complete.  D: the solution added in excess.

2. Match list A with items in list B

    List A.                 
  i)Dropper.       
 ii)Burette.           
 iii)Funnel.
iv)Wash bottle
 v)Pipette
 vi)Retort stand
 vii)Conical flask
viii)White tiles
  ix)Measuring flask
  x)Weghing bottle

  A: measures volume of solutions especially base.

  B: Used to transfer and measure volume of solutions especially acids.

  C: measures accurate volume of liquid solutions.

  D: Add liquids especially indicators in form of drops.

  E:Transfers liquids into containers.

  F: Used as container when measuring mass of solutes.

  G: Holds burette during titration.

  H: Holds pipette during titration.

  I:  Gives clear back ground for observation.

  J:  Keeps distilled water.

  K: Used to mix acid and base when titrating.
  L: Measure specific volume when preparing standard solution.

3. Give the meaning of the following words(statements).
   a) Alkali
   b) Strong base
   c) Molarity
   d) Standardisation
   e) Standard solution
   f) Weak acid
   g) Neutralisation
   h) weak base
   i) Strong acid.

4. During titration methyl orange and phenolphthalein indicators used, state the suitable indicator when titrating.

   i) strong acid against strong base

   ii) weak acid against weak base

   iii) strong base against weak acid

   iv) weak base against strong acid

5. Why it is adviced to add acid into water but not water into a very concentrated acid?

6. Calculate mass of solid potassium chloride to prepare 250cm3 of 0.235M solution.

7. A solution of HCl contains 3.6g of the acid in 1dm3 of solution. 20cm3 of this solution is neutralized by sodium carbonate solution in which 1.1g of it dissolved to make 100cm3 of solution.

(a) Calculate mole ratio for the reaction.

 (b)Calculate the molarities of acid and carbonate solution.

(c) Calculate the volume of the carbonate solution used.

8. 14g sample of impure KOH was dissolved in water to make 100cm3 of solution. The solution was then titrated with 1.26M of nitric acid solution. 40cm3 of the acid  required 25cm3 of alkal solution to be neutralized. Work out the percentage purity of KOH sample.

(b) Which solution seem to be more concentrated for that neutralization reaction? State reason.

9. Washing soda has a formula Na2CO3•H2O
   7.15g of hydrated sodium carbonate dissolved in water to make 250cm3 of its solution. 25cm3 of this solution was titrated with 0.25M of hydrochloric acid. The end point reached when exactly 20cm3 of acid was used.

  Na2CO3 + 2HCl -----> 2NaCl + H2O + CO2.

(a) How the end point for the reaction was observed?

(b) What is the number of moles of the acid used?

(c) Calculate the concentration of the sodium carbonate in mole per litre.

(d) Find the formula molecular mass of hydrated sodium carbonate.

(e) Find molecules of water o crystallization of carbonate compound.

10. Samples of sodium hydroxide solutions of unknown concentration  were titrated against  0.2M of ethanoic acid and the following results obtained.
The volume of pipette used was 10cm3.

(a)  What is the average volume of the acid used

     (b) Calculate molarity of NaOH and its concentration in g/dm3 if the chemical equation is

 NaOH + CH3OOH--->CH3COONa + H2O

   (c) Name the suitable indicator for this reaction and state reason.

   (d) State change of colour observed during titration.

11.   48g of metal carbonate(Y2CO3) were dissolved to make 200cm3 of solution. The solution was titrated with 1M of sulphuric acid, 20cm3 of the solution reacted completely with 37.5cm3 of the sulphuric acid . Determine
   (a) the number of moles of the sulphuric acid used in the titration.
   (b) the number of moles in 20cm3 of the metal carbonate solution.
   (c) the R.A.M of the metal in metal carbonate.

3. a) Refers to the base that dissolves in water.

   b)  Is a base that dissociates completely to form ions.

      *more hydroxide ions formed when strong base dissociated(ionize)

   c) Refers to number of moles of solute dissolved in a solvent to make one litre of solution.

   d) Is a process of determining exact concentration of solution.

   e) Is a solution which its concentration is know.

      *amount of solute and solvent are known in standard solution.

   f) Is an acid that dissociates partially to form ions.

  *few hydrogen ions formed when weak acid ionize.

   g) Is a quantitative reaction between acid and base such than no excess hydrogen or hydroxide ions.

   h)  Is a base that dissociates partially to form ions.

   i) Is an acid that dissociates completely to form ions.

  iii) Phenolphthalein(POP)

5. Because when water added in a very concentrated acid, the solution boils very violently splashing acid out of container.

But when acid added into water no boiling and splashing occur.

    Concentrated acids are denser than water. Adding water into acid forms layer of boiling aced and splashing occurs.

      Adding acid into water, acid flow down and mixes much better with water. No boiling and splashing although the reaction remains exothermic.

      Given data: Molarity = 0.235M

                Relative atomic masses: K=39

     0.235 is number of moles in 1 litre.

           0.235 -----------> 1000cm3

    0.05875  is number of moles of KCl in 250cm3.

    Number of moles(n) = mass/molar mass.

       Mass=n x number of moles

7.(a)The mole ratio for the reaction obtained from the balanced chemical equation

   2HCl + Na2Co3 ------> 2NaCl + Co2 +H2O

The chemical equation shows that:-

     There are two HCl for every 1 NaCo3.

              Molar mass of HCl= (1+35.5)

 Molarity=Concentration(g/dm3)molar mass

Molarity of carbonate solution.

    Its molar mass=[(23 x 2) +12 + (16 x 3)

        Molarity=conc/molar mass

But concentration must be in g/dm3

Concentration of sodium solution is 11g/dm3

        Molarity of acid(Ma)=0.1M

        Molarity of base(Mb)=0.1M

        Volume of acid(Va)=20cm3

        Volume of base(Vb)= required

        Number of moles of acid(na)=2

        Number of moles of base(nb)=1

    Volume(KOH) solution=100cm3

    Volume of acid used=40cm3

    Volume of alkali used=25cm3

    %purity of KOH=required.

In order to get %purity, mass of pure substance and impure substance must obtained.

Mass of pure substance obtained from molarity that we get after titration.

   Mole ratio(na and nb) obtained from the balanced chemical equation.

   KOH + HNO3 ----> KNO3 + H2O

   2M(KOH) is a molarity of KOH made by pure sample of KOH.

 Mass(concentration)=molarity x molar mass

  112g is mass of pure KOH in 1000cm3

Mass of impure KOH is 14g in 100cm3

   140g is mass of impure KOH in 1000cm3.

 Mass of pure KOH/Mass of impure KOH x100%

(b) KOH was more concentrated because small amount of its volume used to neutralise acidic solution. It means the number of hydroxide ions in 25cm3 is equal to the number of hydrogen ions that found in 40cm of acidic solution.

9. (a) The end point for the reaction was observed by colour change of methyl orange from yellow to pink.

    Number of moles in 1000cm3=0.25moles

Number of moles of volume used(20cm3)=?

         ? <---------------- 20cm3="" p="">
The number of moles of acid used =0.005mol.

        Molarity of acid = 0.25M

        Volume of acid used=20cm3

       Molarity of base=required

       Volume of base used=25cm3

      Number of moles of acid=2

      Number of moles of base(n)=1

         Molarity(Na2CO3•H2O)=0.1M

         Its concentration = 7.15g in 250cm3

         Molar mass of Na2CO3•H2O=required

 In order to obtain molar mass. Concentration in g/L have to be determined.

   Molarity = concentration/molar mass.

Molar mass=concentration/molarity

         The molar mass(Na2CO3•H2O)=286g

         Number of molecules of water of

                   crystallization=required.

     (23 x 2) + 12 + (16x 3) + w((1x2)+16)=286

    w  stands for # of water of crystallization

 The number of molecules of water of crytall

   Average volume of the acid used obtained by taking sum of titre values of experiment 1, 2 and 3 divide by number of experiments.

   Titre value obtained by finding different between final burette reading and initial burette reading.

 Titre value for experiment:- 1  = 45.05-21.50

                           experiment  2 =30.00-10.05

                          experiment 3  =40.00-20.00

   Number of experiments is 3

Average volume(acid) = (20.05+19.95+20)/3

        Molarity of acid(Ma)=0.2M

        Volume of acid(Va)= 20cm3

        Molarity of NaOH(Mb)=required

        Volume of base (Vb)=10cm3

        Number of moles of acid(na)=1

        Number of moles of base(nb)=1

    (c) Suitable indicator is Phenolphthalein because the titration is between weak acid against strong base.

    (d) The colour change was from pink to colourless

                 Molarity of sulphuric acid = 1M

                 Number of moles(n) in 37.5cm3= required

                 Volume of sulphuric acid used=37.5cm3

       Molarity means number of moles of the substance which dissolved to make 1 litre of the solution.

   There fore 1M of sulphuric acid meana 1 mole of sulphuric acid found in 1 litre of the solution.

     1 mol -------> 1000cm3(1L)

      0.0375mol used in the reaction.

      (b)  The molarity of the metal carbonate(base) should be calculated first.

            Molarity of acid(Ma) =1M

            Molarity of base(Mb) = required

            Volume of acid = 37.5cm3

    H2SO4  +  Y2CO3 -----> Y2SO4 +  H2O +  CO2

   Number of moles of acid(na)=1

   Numner of moles of base(nb)=1

            1.875mol of metal carbonate dissolved to make 1 litre.

        1.875mol -------> 1000cm3

    0.0375mol found in 20cm3 of metal carbonate solution.

   (c) Concentration in g/litre should be calculated first.

                Molarity = concentration/molar mass

        Whereby molarity = 1.875M

                        Molar mass = (2Y+ 12 + (16 x 3))

       Concentration should be:-

      Molarity = concentration/molar mass

      Molar mass= concentration/molarity

FORM III TOPIC 4. MOLE CONCEPT AND RELATED CALCULATIONS: BANK OF QUESTIONS WITH SOLUTIONS

CHEMISTRY FORM THREE QUESTIONS AND ANSWERS 
FORM III TOPIC 4. MOLE CONCEPT AND RELATED CALCULATIONS: BANK OF QUESTIONS WITH SOLUTIONS

QUESTIONS
1. Multiple choice.

    Choose the most correct answer from the given alternatives.

(i)The mass of one molecule of a compound compared with the mass of one atom of carbon 12 is called _______

  A: molar mass

  B: relative atomic mass

  C: moles

  D: molarity

(ii) Molar mass of 2H2O (water) ...............

  A: 18    B: 36   C: 17   D: 54

(iii) .............. is a ratio of mass or volume of the substance to its molar mass or molar volume.

  A: mole

  B: molarity

  C: concentration

  D: relative atomic mass

(iv) In order to obtain ................ , Avogadro's constant is multiplied with number of moles of the substance.

 A: molar mass

 B: molar solution

 C: electrons

 D: number of particles.

(v) .................. is any known amount of solute dissolved in water to make one litre.

  A: molar solution

  B: concentration

  C: molarity

  D: number of particles

(vi) The relationship of quantities between reactants and products in a chemical reaction is known as...............

  A: stoichiometry

  B: quantitative ratio

  C: Avogadro's number

  D: mole ratio

(vii) The mass of one mole of the substance is known as

 A: molarity

 B: mass

 C: molar mass

 D: molar solution

(viii) One mole of solute forms ................ when dissolved to make one litre of the solution.

   A: molarity

   B: molar solution

   C: molar volume

   D: moles

 (ix) A number of atoms in exactly 12g of carbon 12 isotope

   A: number of particles

   B: Avogadro's number

   C: number of moles

   D: number of molecules

(x) At standard temperature and pressure equal volume of all gases contain same number of particles in ..................

  A: 22.4 L

  B: 22.4cm3

  C: 1 L

  D. 1000cm3


2. Matching items.

    Match items in list A with responses in list B

       List A

(i) concentration in mole per litre

(ii) number of particles of 20g of Ca

(iii) volume occupied by one mole of gas in 22.4dm3 at s.t.p

(iv)molarity of 10g of NaOH in 500cm3

(v) Avogadro's constant


         List B

A: molar volume

B: 6.02 x 10^23

C: molarity

D:  3.01 x 10^-23

E: molarity

F: 0.5


3. Define the terms

  (a) mole

  (b) molar mass

  (c) molarity

  (d) concentration

  (e) molar volume

  (f) molar solution.


4. (a) What volume of hydrogen gas is produced at STP when  0.65g of zinc reacts completely with excess hydrochloric acid?

     (b) What mass of sulphuric acid  is found in 400 cm^3 of its 0.1M aqueous solution?

    (c) What is the molarity of :-

   (i)solution containing 5.3g of sodium carbonate in 100ml of the solution?

   (ii) solution containing 80g of sodium hydroxide in 2 litre?


5. Calculate amount of substance in the following:-

    (a) 0.69g of sodium

    (b) 180g of carbon

    (c) 0.98g of sulphuric acid


6.(i) Find the mass of the following substances

   (a) 0.78 mol of Ca(CN)2

   (b) 7 mol of hydrogen peroxide

   (c) 3.01 x 10^-23 particles of Ca.

  (ii) Find the number of particles in the following substances

      (a) 16.8 litres of chlorine gas at STP.

     (b) 0.8 mol of nitric acid.

     (c) 160g sodium carbonate

7. A gas occupies 1.95dm^3 at STP. Calculate the number of :-

      (i) moles in the gas

      (ii) molecules in the gas.


8.(a) Ammonium chloride reacts with calcium hydroxide as shown in the following equation

2NH4Cl (aq)  +  Ca(OH)2 (aq) ----->

  2NH3 (g)  +  CaCl2 (aq)  + 2H2O(l)

 Calculate the mass of ammonium chloride that will react completely with 14.8g of calcium hydroxide.

(b) Sodium sulphate reacts with barium nitrate as shown in the equation below.

Na2So4 (aq)  +  Ba(NO3)2  --------->

     BaSo4 (s)  +  2NaNO3 (aq)

How many moles of barium sulphate are produced when 80.5g of sodium sulphate react completely with barium nitrate?


9. (a) What is the relative molecular mass of gas Y if the mass of 560 cm^3 of gas Y measured at STP is 1.1g? (Y is not the actual symbol of the gas)

  (b) 4.9g of copper (II) oxide yield 3.92g of copper when reduced by hydrogen gas. Calculate the mass of off copper (II) oxide required to yield 4g of copper. (Cu=64, O=16)


10. (a) 20cm^3 of NaOH solution containing 8gdm^3 was required for complete neutralization of 0.18g of dibasic acid. Calculate the relative molecular mass of the acid(Na=23, O=16, H=1).

     (b) What mass of CO2 (g) will occupy 224cm^3 at STP (C=12, O=16


11.🏾What is the concentration of resulting solution of HCl, when 100cm3 of 0.1M HCl mixed with 100cm3 of 1M HCl?



        SOLUTIONS.

1. (i) C   (ii) A   (iii) A  (iv) D   (v) C

  (vi) A  (vii) C  (viii) B  (ix) B   (x)  A


2. (i) C   (ii) D   (iii) A   (iv) F   (v) B


3.(a) Mole of substance is the amount of a substance which contains the Avogadro’s number of particles.

   (b) Molar mass:- is a mass of one mole of the compound.

     For example. 2H2SO4 (sulphuric acid)

-its molar mass = Mole of substance is the amount of

a substance which contains the Avogadro’s

number of particles.

  -its composition by mass

      = 2((1x2)+32+(16x4))

       = 196g.

 (c) Molarity :- is a number of moles of moles of solute in one litre of the solution.

  (d)molar volume:- is number of moles of gas in 22.4L at standard temperature and pressure.

(e) molar solution :- is a solution which contains one mole of solute dissolved in a solvent to make one litre(1dm^3) of the solution.


4. (a) given data.

     Mass = 0.65g

     Molar mass = 98g/mol.

     Volume = needed

Firstly calculate number of moles then convert into volume

     n= mass

          Molar mass

       = 0.65

          65.5

       = 0.0099mol.

From the eqn.

 Zn  +  2HCl ------> ZnCl2  + H2

This means

  1mol.(Zn) -------> 1mol. (H)

   0.0099  ----------> ?

  ? = 0.0099 x 1

               1

   = 0.0099mol(H)

 Convert the obtained moles of hydrogen into volume.

  From.

       n = volume

              Molar gas volume

   Volume = n x molar gas volume

                  = 0.0099 x 22.4

                  = 0.222dm^3.

 (b) given data.

    Molarity = 0.1M

    Volume = 400cm^3

     Mass  = needed

      Molar mass = 98.

Firstly find the number of moles in 400cm^3 then convert moles into mass.

Remember that Molarity means the number of moles of solute that dissolved to make one litre of the solution.

    0.1mol. -------> 1000cm^3

           ? <----------- 400="" cm="" p="">
          ?= 0.1 x 400

                 1000

          0.04mol(in 400cm^3)

Convert 0.04mol. Into mass.

      n = mass.

            Molar mass

  Mass = n x molar mass

             = 0.04 x 98

             = 3.92g.

(c) (i) data given.

      Volume = 100cm^3

      Mass = 5.3g

      Molarity = required

       Molar mass = 106g/mol.

But molarity = number of mole in one litre of the solution.

  Also = concentration(g/dm^3)

              Molar mass

Convert 5.3g in 100cm^3 into g/L(litre)

 5.3g ---------> 100cm^3

     ? <----------- 1000cm="" p="">
   ? = 5.3 x 1000

              100

        = 53g/dm^3

 Molarity = concentration

                     Molar mass

              = 53

                 106

             = 0.5M

 (ii)

 data given.

      Volume = 2L (2000cm^3)

      Mass = 80g

      Molarity = required

       Molar mass = 40g/mol.

But molarity = number of mole in one litre of the solution.

  Also = concentration(g/dm^3)

              Molar mass

Convert 80g in 2000cm^3 into g/L(litre) or dm^3

 80g ---------> 2000cm^3

     ? <----------- 1000cm="" p="">
   ? = 80 x 1000

              2000

        = 40g/dm^3

 Molarity = concentration

                     Molar mass

              = 40

                 40

             = 1M


5. (a) Given data.

     Mass = 0.69g

    Molar mass = 23g/mol.

    Number of moles(n) = needed

     n = mass

           Molar mass

     = 0.69
          23

    = 0.03mol.

(b) given data.

     Mass = 180g

    Molar mass = 12g/mol.

    Number of moles(n)=required

     n = Mass/ Molar mass

      = 180
          12

        = 15mol.

 (c)  Mass = 0.98g

    Molar mass = 98g/mol.

    Number of moles(n)=required

     n = mass /Molar mass

     = 0.98/ 98
     =o.1mol.


6. (i) a) given data.
    Number of mol. = 0.78mol.
   Molar mass(Ca(CN)2) = 92g/mol
  Mass of (Ca(CN)2) = needed
   From.
       n=     mass
            Molar mass
  Mass = n x molar mass
             = 0.78 x 92
             = 71.76g
 (b) given data.
       n (H2O2) = 7mol.
      Molar mass (H2O2) = 34g/mol
       mass (H2O2) = needed
      Mass = n x molar mass
                 = 7 x 34
                 = 34g/mol.
  (c) given data.
Number of particles(N)
                 =3.01 x 10^23
 LA• = 6.02 x 10^23
 Mass= needed
 Firstly find the number of moles (n)
 From.
      N = n x LA•
     n= N
           LA•
         = 3.01 x 10^23
             6.02 x 10^23
         = 0.5
  Mass = n x molar mass
             = 0.5 x 40
             = 20g/mol.
  (ii) (a) given data.
         Volume = 16.8L
         Molar gas volume = 22.4L
         Number of particles(N) = needed
      Firstly calculate number of moles.
     From.
          n = volume
                Molar volume
              = 16.8
                  22.4
                = 0.75mol.
Where by
        N= n x LA•
           = 0.75 x 6.02 x 10^23
           =4.515 particles
  (b) given data.
       n = 0.8mol.
       LA• = 6.02 x 10^23
        N  = needed
From.
      N= n x LA•
         = 0.8 x 6.02 x 10^23
         = 4.816 particles
(c) given data.
      Mass = 160g
      Molar mass = 106g/mol.
       N= needed
      LA• = 6.02.x 10^23
   N= n x LA•
Whereby
       n = mass
            Molar mass
          = 160
              106
           = 1.509mol.
   N= 1.509 x 6.02 x 10^23
      = 9.084 particles

5.

7. (i) given data.
     Volume of gas = 1.95dm^3
     Molar gas volume = 22.4dm^3
     Number of moles = needed
     From.
         n =      Volume
               Molar gas volume
            = 1.95
               22.4
           = 0.087mol.
   (ii) Available data.
        Number of moles(n) = 0.087mol.
        Avogadro's constant (LA•) = 6.02 x 10^23
  Number of moles = needed
  Number o particles = n x LA•
            = 0.087 x 6.02 x 10^23
            = 0.524 x 10^23 particles.

8. (a) Given data.
         Mass(NH4Cl) = needed
        Mass((Ca(OH)2) = 14.8
From the eqn.
 2NH4Cl  +  Ca(OH)2 ---------> 2NH3 + CaCl2 + 2H2O
 Mass(NH4) = 2(14 + (1 x 4))
                      = 36g
Mass(Ca(OH)2) = (40 + 2(16 + 1)
                             = 74g
   36g(NH4) --------> 74g (Ca(OH)2)
               ?<----------- 14.8="" p="">         ?= 36 x 14.8
                   74
                 =7.2g
     7.2g of NH4 were needed
Alternative solution.
Number of moles can be used then converted into mass.
Firstly convert 14.8g of Ca(OH)2 into moles.
    n =    mass
          Molar mass
       = 14.8
          74
       0.2mol.
Equation shows that:-
  2mol.(NH4) ----------> 1mol.(Ca(OH)4)
                ? <------------- 0.2mol="" nbsp="" p="">         ? = 2 x 0.2
                   1
          =0.4 mol(NH4)
     Convert it into mass.
     Mass = n x molar mass
     Molar mass of ammonium = 18
    Mass of ammonium needed
                  0.4 x 18
                = 7.2g
 (b) data given.
   Mass(Na2SO4) =80.5g
   From the eqn.
  Na2SO4 + Ba(NO3)2 -------> NaNO3 + BaSO4
   It shows that:-
 Mass(Na2SO4) = 142g
 Mass (Ba(NO3)2 = 261.3g
142g(Na2SO4)-->261.3g(Ba(NO3)2
 80.5g       -----------> ?
   ? = 80.5 x 261.3
            142
      = 148.131g Ba(NO3)2  required
     Convert nto moles.
      n =    mass
            Molar mass
          = 148.131
              261.3
         = 0.567mol.
 Alternative solution.
Firstly convert 80.5g(Na2SO4) into moles.
     n = 80.5
            142
         = 0.567mol.
Eqn. Shows that
 1mol (Na2SO4 ---->1mol(Ba(NO3)2
 0.567           ----------> ?
         = 0.567mol.(Ba(NO3)2

9. (a) given data.
      Volume of gas Y at STP= 560 cm^3
    Mass of gas Y at STP = 1.1g

Calculate number of moles then use number of moles obtained to find the molar mass of the gas since its mass is given.
Fro n= volume (dm^3)
            Molar gas volume(dm^3)
 (b)  data given.
 Mass of copper oxide = 4.9g
 Mass if copper  = 3.9
Mass of copper oxide = needed, when 4g of copper produced.
  CuO  + H2 -----> Cu  +  H2O
  4.9g(CuO) ---------> 3.92(Cu)
                ? <----------- 4g="" p="">          ? = 4.9 x  4
                   3.92
             =5g
   5g of CuO were needed
10. Given data
    Volume of NaOH used=20
    Concentration NaOH = 8g/dm^3
    Mass of the acid used = 0.18g
    Molar mass of acid = needed.
From balanced chemical equation
 2NaOH   + H2X -----> Na2X  + 2H2O
Base acid mole ratio is 2:1
 This means base volume used is twice of acid used.
There fore in order to get volume of acid, volume of base must be divided by 2
  Volume of acid = 20cm^3
                                      2
                              =10 cm^3
 Now find the molarity of acid base then molarity if acid.
 From
   Molarity(NaOH) = conc.
                               Molar mass
                                    8   
                                   40
                             0.2M
From
  MaVa  = na
  MbVb     nb
    Whereby Mb = 0.2M
                     Ma =needed
                     Vb  = 20cm^3
                     Va  = 10cm^3
                     na  = 1
                     nb  = 2

      Ma x 10  = 1
      0.2 x 20     2
      Ma = 0.2
 Find concentration (g/dm^3)

   Concentration = mass(g)
                              Volume(dm^3)
   .convert 10cm^3 into dm^3 but the mass of acid is 0.18g.
        Concentration = 0.18g
                                      0.01dm^3
                        =18g/dm^3
Since Molarity = concentration
                                Molar mass

  Molar mass = concentration
                               Molarity
          = 18
             0.2
          =90
 Relative mass of the acid is 90g/mol
 Alternative solution.
   Given data.
    Vb=20cm^3 (0.02dm^3)
    Concentration of base(b) = 8g/dm^3
  Mass of acid  = 0.18g
 Mb=concentration
           Molar mass
                      = 8
                        40
                     = 0.2M

From.
     MaVa = na
     MbVb    nb
 Mava = MbVbna
                    nb

            = 0.2 x 20 x 1
                    2
MaVa = 0.002
MaVa = na
There fore number of moles of acid=0.002
 From n= mass
               Molar mass
  Molar mass = mass
                            n(number of moles)
                        = 0.18
                           0.002
                        = 90g/mol.

11. Data given:
    Molarities of HCl = 0.1M and 1M
    Volumes=both have 100cm3

   Calculate the number of moles of solutions in 100cm3 per each solution because 0.1M and 1M are number of moles in 1 litre in each solution.
* number of moles of 0.1M HCl in 100cm3
    0.1mol ---------> 1000cm3
         ?  <----------- 100cm3="" p="">            = 0.01mol.
 *number of moles of 1M HCl in 100cm3
    1mol ---------> 1000cm3
         ?  <----------- 100cm3="" p="">            = 0.1mol.
Sum up the number of moles obtained then find the concentration by using total number of moles and total volume.
 Total number of mol=0.01+ 0.1
                                      =0.11mol.
Total volume= 100cm^2 + 100cm^3
                       = 200cm^3 = 0.2dm^3
   From.
   Concentration =   Number of moles
                                    Volume(dm3)
                             = 0.11
                                 0.2
                            = 0.55M