WAEC SSCE - Chemistry - 2019 (Essay)
Ajụjụ 1 Ripọtì
(a) Explain briefly why \(^{4}_{2}\mathrm{He}\) has a stable electron configuration compared to \(^{9}_{4}\mathrm{Be}\)
(b) Consider the following elements: 1H and 3Li.
(i) State the number of electrons that an atom of each element would have after forming an ionic bond.
(ii) Give a reason for each of your answers stated in (b)(i).
(c) State two factors that should be considered when siting a chemical industry.
(d) State two advantages of using a catalyst instead of high temperatures in chemical reactions.
(e) Turpentine burns in chlorine according to the following equation:
\[ \mathrm{C}_{10}\mathrm{H}_{16(l)} + 8\mathrm{Cl}_{2(g)} \to 10\mathrm{C}_{(s)} + 16\mathrm{HCl}_{(g)} \]
Calculate the mass of turpentine that would completely burn in 21.3 g of chlorine.
[Molar mass of chlorine = \(71\ \mathrm{gmol}^{-1}\); Molar mass of Turpentine = \(136\ \mathrm{gmol}^{-1}\)]
(f) What is cracking?
(g) State two factors that may influence the value of electron affinity.
(h) What are carbohydrates?
(i) State two differences between a simple sugar and starch.
(j) Write an equation to show the dissociation of each of the following acids:
(i) \(\mathrm{H}_2\mathrm{CO}_3\);
(ii) \(\mathrm{CH}_3\mathrm{COOH}\).
(a) Helium has the electron configuration 2 (a completely filled first shell, a stable duplet), so its outermost shell is full and it has no tendency to gain or lose electrons. Beryllium is 2,2; its outermost shell holds only 2 of the electrons needed to complete it, so it is not a stable arrangement and beryllium readily loses those electrons.
(b)(i) In forming an ionic bond (as in LiH), hydrogen gains one electron to become H- with 2 electrons, and lithium loses one electron to become Li+ with 2 electrons.
(ii) Hydrogen (1 electron) gains one electron to attain the stable duplet of helium; lithium (2,1) loses its single outer electron to attain the stable duplet of helium.
(c) Any two: availability of raw materials; availability of water; nearness to market; good transport network; availability of cheap labour and power.
(d) A catalyst lowers energy cost because a high temperature is not needed; and it is not used up (it can be reused), and it avoids decomposition of heat-sensitive products.
(e) \(C_{10}H_{16} + 8Cl_2 \rightarrow 10C + 16HCl\)
Amount of chlorine \(= \dfrac{21.3}{71} = 0.300\ \text{mol}\).
Mole ratio turpentine : chlorine \(= 1 : 8\), so amount of turpentine \(= \dfrac{0.300}{8} = 0.0375\ \text{mol}\).
Mass of turpentine \(= 0.0375 \times 136 = \mathbf{5.10\ g}\).
(f) Cracking is the breaking down of large (long-chain) hydrocarbon molecules into smaller, more useful molecules using heat and/or a catalyst.
(g) Any two: atomic size (radius); nuclear charge; shielding (screening) effect of inner electrons; existing electron configuration.
(h) Carbohydrates are organic compounds of carbon, hydrogen and oxygen in which hydrogen and oxygen are present in the ratio 2:1 (general formula \(C_x(H_2O)_y\)); they are polyhydroxy aldehydes or ketones.
(i) Any two: a simple sugar is soluble in water while starch is insoluble; a simple sugar is sweet while starch is not; a simple sugar is a small molecule while starch is a large polymer (macromolecule); a simple sugar reduces Fehling's solution while starch does not.
(j)(i) \(H_2CO_3 \rightleftharpoons 2H^+ + CO_3^{2-}\)
(ii) \(CH_3COOH \rightleftharpoons CH_3COO^- + H^+\)
Akọwa Nkọwa
(a) Helium has the electron configuration 2 (a completely filled first shell, a stable duplet), so its outermost shell is full and it has no tendency to gain or lose electrons. Beryllium is 2,2; its outermost shell holds only 2 of the electrons needed to complete it, so it is not a stable arrangement and beryllium readily loses those electrons.
(b)(i) In forming an ionic bond (as in LiH), hydrogen gains one electron to become H- with 2 electrons, and lithium loses one electron to become Li+ with 2 electrons.
(ii) Hydrogen (1 electron) gains one electron to attain the stable duplet of helium; lithium (2,1) loses its single outer electron to attain the stable duplet of helium.
(c) Any two: availability of raw materials; availability of water; nearness to market; good transport network; availability of cheap labour and power.
(d) A catalyst lowers energy cost because a high temperature is not needed; and it is not used up (it can be reused), and it avoids decomposition of heat-sensitive products.
(e) \(C_{10}H_{16} + 8Cl_2 \rightarrow 10C + 16HCl\)
Amount of chlorine \(= \dfrac{21.3}{71} = 0.300\ \text{mol}\).
Mole ratio turpentine : chlorine \(= 1 : 8\), so amount of turpentine \(= \dfrac{0.300}{8} = 0.0375\ \text{mol}\).
Mass of turpentine \(= 0.0375 \times 136 = \mathbf{5.10\ g}\).
(f) Cracking is the breaking down of large (long-chain) hydrocarbon molecules into smaller, more useful molecules using heat and/or a catalyst.
(g) Any two: atomic size (radius); nuclear charge; shielding (screening) effect of inner electrons; existing electron configuration.
(h) Carbohydrates are organic compounds of carbon, hydrogen and oxygen in which hydrogen and oxygen are present in the ratio 2:1 (general formula \(C_x(H_2O)_y\)); they are polyhydroxy aldehydes or ketones.
(i) Any two: a simple sugar is soluble in water while starch is insoluble; a simple sugar is sweet while starch is not; a simple sugar is a small molecule while starch is a large polymer (macromolecule); a simple sugar reduces Fehling's solution while starch does not.
(j)(i) \(H_2CO_3 \rightleftharpoons 2H^+ + CO_3^{2-}\)
(ii) \(CH_3COOH \rightleftharpoons CH_3COO^- + H^+\)
Ajụjụ 2 Ripọtì
(a) A hydrocarbon having the formula \( \mathrm{C}_{10}\mathrm{H}_{22} \) was cracked to produce \( \mathrm{C}_6\mathrm{H}_{14} \) and another hydrocarbon P.
(i) Give the molecular formula of P.
(ii) Draw the structures of two isomers of P.
(ii) Give a reason why P could be polymerized.
(b) State the guiding principles which are used to explain the way electrons of the atoms of the elements are arranged in atomic orbitals.
(c) Consider each of the following substances: NaH, \( \mathrm{H}_2 \), \( \mathrm{H}_2\mathrm{S} \), \( \mathrm{NH}_4\mathrm{Cl} \).
(i) Describe the nature of the intermolecular forces holding the units or molecules together in the condensed (liquid or solid) state.
(ii) Explain briefly what happens when a sample of each of the substances is added to water.
(iii) Write the chemical equations of any reactions occurring or of any equilibria established.
(d) Element J has the following electron configuration: \( 1\mathrm{s}^2 2\mathrm{s}^2 2\mathrm{p}^6 3\mathrm{s}^2 \).
(i) How many unpaired electrons can be found in J?
(ii) State whether J would be a good oxidizing or reducing agent.
(iii) Give a reason for the answer in (d)(ii).
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Cracking of a Hydrocarbon
A hydrocarbon with formula C10H22 is cracked to produce C6H14 and another hydrocarbon, P.
(a)(i) Molecular formula of P
C10H22 → C6H14 + P
P = C10H22 − C6H14 = C4H8
Therefore, P is an alkene with the molecular formula C4H8.
(a)(ii) Two isomers of P
The structures shown are but-1-ene, CH2=CHCH2CH3, and 2-methylpropene, CH2=C(CH3)2.
(a)(iii) Why P can be polymerised
P can be polymerised because it is an alkene containing a carbon–carbon double bond (C=C). The double bond can open and form bonds with other P molecules, producing a long-chain polymer.
Akọwa Nkọwa
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Cracking of a Hydrocarbon
A hydrocarbon with formula C10H22 is cracked to produce C6H14 and another hydrocarbon, P.
(a)(i) Molecular formula of P
C10H22 → C6H14 + P
P = C10H22 − C6H14 = C4H8
Therefore, P is an alkene with the molecular formula C4H8.
(a)(ii) Two isomers of P
The structures shown are but-1-ene, CH2=CHCH2CH3, and 2-methylpropene, CH2=C(CH3)2.
(a)(iii) Why P can be polymerised
P can be polymerised because it is an alkene containing a carbon–carbon double bond (C=C). The double bond can open and form bonds with other P molecules, producing a long-chain polymer.
Ajụjụ 3 Ripọtì
(a) In the Solvay process, explain briefly with equations the functions of the following substances:
- limestone;
- ammonia.
(b) (i) Write a chemical equation for the fermentation of glucose.
(ii) Explain briefly why a tightly-corked glass bottle filled to the brim with fresh palm-wine shatters on standing for some time.
(c) Consider the following metals: Na, Fe, K and Cu.
(i) Arrange the metals in order of increasing reactivity.
(ii) Which of the metals will react with cold water?
(ii) Which of the metals could form coloured salts?
(d)(i) What is a redox reaction?
(ii) Identify which of the following reaction equations are redox.
(I) \(2\mathrm{Na} + \mathrm{Cl}_2 \rightarrow 2\mathrm{NaCl}\)
(II) \(\mathrm{AgCl} + 2\mathrm{NH}_3 \rightarrow [\mathrm{Ag}(\mathrm{NH}_3)_2]\mathrm{Cl}\)
(III) \(\mathrm{C}_2\mathrm{H}_2 + \mathrm{H}_2 \rightarrow \mathrm{C}_2\mathrm{H}_4\)
(IV) \(\mathrm{HCl} + \mathrm{KOH} \rightarrow \mathrm{KCl} + \mathrm{H}_2\mathrm{O}\)
(V) \(2\mathrm{FeCl}_3 + 2\mathrm{KI} \rightarrow 2\mathrm{FeCl}_2 + 2\mathrm{KCl} + \mathrm{I}_2\)
(iii) Give a reason for each of the answers in (d)(ii).
(iv) Write balanced equations of the half reactions for any two of the redox reactions in (d)(ii).
Solvay Process and Chemical Reactions
(a) Functions in the Solvay Process:
- Limestone (CaCO3):
- Function: Source of calcium oxide (CaO) for the regeneration of ammonia.
- Equation: CaCO3 (g) -> CaO (s) + CO2 (g) (Limestone decomposes to lime and carbon dioxide)
- Ammonia (NH3):
- Function:
- Reacts with calcium and chloride ions in brine to form ammonium chloride (NH4Cl).
- Makes the solution alkaline, favoring the precipitation of calcium carbonate.
- Equation: Not directly involved in a reaction, but participates in the overall process.
- Function:
(b) Fermentation and Shattered Bottle:
(i) Glucose Fermentation:
C6H12O6 (aq) -> 2C2H5OH (aq) + 2CO2 (g) (Glucose ferments to ethanol and carbon dioxide)
(ii) Shattered Bottle:
Fresh palm wine undergoes fermentation by yeast. During fermentation, yeast consumes sugar and produces carbon dioxide gas. In a tightly sealed bottle, the CO2 gas gets trapped and builds up pressure. As pressure increases, the glass bottle can no longer withstand the force and shatters.
(c) Metal Reactivity:
(i) Increasing Reactivity: K > Na > Fe > Cu
(ii) Reacting with Cold Water: K, Na (These alkali metals react vigorously with water)
(iii) Colored Salts: Cu (Copper forms characteristic colored salts, like blue copper sulfate)
(d) Redox Reactions and Half Reactions:
(i) Redox Reaction:
A redox reaction is a type of chemical reaction where there is a transfer of electrons between the participating atoms or ions. Oxidation is the loss of electrons, and reduction is the gain of electrons.
(ii) Identifying Redox Reactions:
- Redox: (II), (V)
- Not Redox: (I), (III), (IV)
(iii) Explanation for Redox Reactions:
- (II) AgCl + 2NH3 -> [Ag(NH3)2]Cl:
- Silver (Ag) loses an electron to become Ag+ (oxidation).
- Nitrogen (N) in ammonia gains an electron to form the complex ion [Ag(NH3)2]+ (reduction).
- (V) 2FeCl3 + 2KI -> 2FeCl2 + 2KCl + I2:
- Iron (Fe) in FeCl3 loses an electron to become Fe2+ in FeCl2 (oxidation).
- Iodine (I) gains an electron to form I2 (reduction).
(iv) Balanced Half Reactions (Examples):
- Oxidation Half Reaction (from reaction II):
Ag -> Ag+ + e-
- Reduction Half Reaction (from reaction V):
I- -> I2 + e- (Note: I- needs to be balanced with Cl- from the reactant side)
Akọwa Nkọwa
Solvay Process and Chemical Reactions
(a) Functions in the Solvay Process:
- Limestone (CaCO3):
- Function: Source of calcium oxide (CaO) for the regeneration of ammonia.
- Equation: CaCO3 (g) -> CaO (s) + CO2 (g) (Limestone decomposes to lime and carbon dioxide)
- Ammonia (NH3):
- Function:
- Reacts with calcium and chloride ions in brine to form ammonium chloride (NH4Cl).
- Makes the solution alkaline, favoring the precipitation of calcium carbonate.
- Equation: Not directly involved in a reaction, but participates in the overall process.
- Function:
(b) Fermentation and Shattered Bottle:
(i) Glucose Fermentation:
C6H12O6 (aq) -> 2C2H5OH (aq) + 2CO2 (g) (Glucose ferments to ethanol and carbon dioxide)
(ii) Shattered Bottle:
Fresh palm wine undergoes fermentation by yeast. During fermentation, yeast consumes sugar and produces carbon dioxide gas. In a tightly sealed bottle, the CO2 gas gets trapped and builds up pressure. As pressure increases, the glass bottle can no longer withstand the force and shatters.
(c) Metal Reactivity:
(i) Increasing Reactivity: K > Na > Fe > Cu
(ii) Reacting with Cold Water: K, Na (These alkali metals react vigorously with water)
(iii) Colored Salts: Cu (Copper forms characteristic colored salts, like blue copper sulfate)
(d) Redox Reactions and Half Reactions:
(i) Redox Reaction:
A redox reaction is a type of chemical reaction where there is a transfer of electrons between the participating atoms or ions. Oxidation is the loss of electrons, and reduction is the gain of electrons.
(ii) Identifying Redox Reactions:
- Redox: (II), (V)
- Not Redox: (I), (III), (IV)
(iii) Explanation for Redox Reactions:
- (II) AgCl + 2NH3 -> [Ag(NH3)2]Cl:
- Silver (Ag) loses an electron to become Ag+ (oxidation).
- Nitrogen (N) in ammonia gains an electron to form the complex ion [Ag(NH3)2]+ (reduction).
- (V) 2FeCl3 + 2KI -> 2FeCl2 + 2KCl + I2:
- Iron (Fe) in FeCl3 loses an electron to become Fe2+ in FeCl2 (oxidation).
- Iodine (I) gains an electron to form I2 (reduction).
(iv) Balanced Half Reactions (Examples):
- Oxidation Half Reaction (from reaction II):
Ag -> Ag+ + e-
- Reduction Half Reaction (from reaction V):
I- -> I2 + e- (Note: I- needs to be balanced with Cl- from the reactant side)
Ajụjụ 4 Ripọtì
(a) State the conditions necessary for the cracking of long-chain hydrocarbons to produce more gasoline.
(b) State two reasons why metallic objects are electroplated.
(c) (i) Explain briefly why calcium oxide cannot be used to dry hydrogen chloride gas.
(ii) State one drying agent for hydrogen chloride gas.
(d) Concentrated trioxonitrate (V) acid was added to a solution of iron (II) tetraoxosulphate (VI) and the mixture heated. The mixture turned from pale green to yellow with the evolution of a brown gas. Explain briefly these observations.
(e) (i) Write the equation for the reaction between zinc oxide and
- dilute tetraoxosulphate (VI) acid,
- sodium hydroxide solution.
(ii) State which property of zinc oxide is shown by the reaction in (e)(i).
(f) Two isotopes of chlorine are \(^{35}_{17}\mathrm{Cl}\) and \(^{37}_{17}\mathrm{Cl}\). State one:
- similarity;
- difference between the isotopes.
(g) State the two products formed when chlorine water is exposed to sunlight.
(h) Consider the reaction represented by the following equation:
State the:
- species that is undergoing oxidation;
- oxidizing agent.
(i) What is meant by carbon-12 scale?
(j) State two properties of a chemical system in equilibrium.
(a) Conditions for cracking long-chain hydrocarbons to give more gasoline
A high temperature (about 400 - 600 °C) together with a catalyst (aluminosilicate / zeolite, i.e. catalytic cracking). Purely thermal cracking uses high temperature and high pressure.
(b) Two reasons metallic objects are electroplated
- To protect the metal from corrosion / rusting.
- To improve the appearance (make it bright and decorative). (Also: to resist wear and increase durability.)
(c)(i) Why calcium oxide cannot dry hydrogen chloride gas
Calcium oxide is a basic oxide, and hydrogen chloride is an acidic gas; the two react chemically (CaO + 2HCl → CaCl2 + H2O) instead of the CaO merely removing moisture, so the gas would be absorbed and lost.
(c)(ii) One drying agent for HCl gas: concentrated tetraoxosulphate(VI) acid (conc. H2SO4). (Anhydrous calcium chloride is also acceptable.)
(d) Conc. HNO3 added to iron(II) tetraoxosulphate(VI) and heated
Concentrated trioxonitrate(V) acid is a strong oxidizing agent. It oxidizes the pale-green Fe2+ ion to the yellow (brown-yellow) Fe3+ ion, which accounts for the colour change from pale green to yellow. In being reduced, the acid liberates nitrogen(IV) oxide, the brown gas observed.
\[3Fe^{2+} + 4H^{+} + NO_3^{-} \longrightarrow 3Fe^{3+} + NO\uparrow + 2H_2O\]
(with concentrated acid the brown gas evolved is NO2).
(e)(i) Zinc oxide with dilute acid and with alkali
\[ZnO + H_2SO_4 \longrightarrow ZnSO_4 + H_2O\]
\[ZnO + 2NaOH \longrightarrow Na_2ZnO_2 + H_2O\]
(sodium zincate is formed with the alkali).
(e)(ii) Property shown: zinc oxide is amphoteric (it reacts with both acids and bases).
(f) Isotopes 3517Cl and 3717Cl
- Similarity: both have the same atomic (proton) number, 17, hence the same number of electrons and the same chemical properties.
- Difference: they have different mass numbers (35 and 37), i.e. different numbers of neutrons (18 and 20).
(g) Two products when chlorine water is exposed to sunlight
Chlorine water (HOCl + HCl) decomposes in sunlight, 2HOCl → 2HCl + O2, giving hydrogen chloride (hydrochloric acid) and oxygen.
(h) For the reaction 2H2S + SO2 → 3S + 2H2O
Assigning oxidation numbers to sulphur: it is −2 in H2S, +4 in SO2, and 0 in the free sulphur produced.
- Species undergoing oxidation: H2S - the sulphur in it rises from −2 to 0 (loses electrons).
- Oxidizing agent: SO2 - the sulphur in it falls from +4 to 0 (it is reduced, so it oxidizes the H2S).
(i) Carbon-12 scale
It is the scale of relative atomic masses on which one atom of the carbon-12 isotope is assigned a mass of exactly 12.000 units, and the mass of every other atom is measured relative to one-twelfth of the mass of a carbon-12 atom.
(j) Two properties of a chemical system in equilibrium
- It is dynamic: the forward and backward reactions continue at equal rates.
- Its macroscopic properties (concentrations, colour, pressure) remain constant, and it is attained only in a closed system.
Akọwa Nkọwa
(a) Conditions for cracking long-chain hydrocarbons to give more gasoline
A high temperature (about 400 - 600 °C) together with a catalyst (aluminosilicate / zeolite, i.e. catalytic cracking). Purely thermal cracking uses high temperature and high pressure.
(b) Two reasons metallic objects are electroplated
- To protect the metal from corrosion / rusting.
- To improve the appearance (make it bright and decorative). (Also: to resist wear and increase durability.)
(c)(i) Why calcium oxide cannot dry hydrogen chloride gas
Calcium oxide is a basic oxide, and hydrogen chloride is an acidic gas; the two react chemically (CaO + 2HCl → CaCl2 + H2O) instead of the CaO merely removing moisture, so the gas would be absorbed and lost.
(c)(ii) One drying agent for HCl gas: concentrated tetraoxosulphate(VI) acid (conc. H2SO4). (Anhydrous calcium chloride is also acceptable.)
(d) Conc. HNO3 added to iron(II) tetraoxosulphate(VI) and heated
Concentrated trioxonitrate(V) acid is a strong oxidizing agent. It oxidizes the pale-green Fe2+ ion to the yellow (brown-yellow) Fe3+ ion, which accounts for the colour change from pale green to yellow. In being reduced, the acid liberates nitrogen(IV) oxide, the brown gas observed.
\[3Fe^{2+} + 4H^{+} + NO_3^{-} \longrightarrow 3Fe^{3+} + NO\uparrow + 2H_2O\]
(with concentrated acid the brown gas evolved is NO2).
(e)(i) Zinc oxide with dilute acid and with alkali
\[ZnO + H_2SO_4 \longrightarrow ZnSO_4 + H_2O\]
\[ZnO + 2NaOH \longrightarrow Na_2ZnO_2 + H_2O\]
(sodium zincate is formed with the alkali).
(e)(ii) Property shown: zinc oxide is amphoteric (it reacts with both acids and bases).
(f) Isotopes 3517Cl and 3717Cl
- Similarity: both have the same atomic (proton) number, 17, hence the same number of electrons and the same chemical properties.
- Difference: they have different mass numbers (35 and 37), i.e. different numbers of neutrons (18 and 20).
(g) Two products when chlorine water is exposed to sunlight
Chlorine water (HOCl + HCl) decomposes in sunlight, 2HOCl → 2HCl + O2, giving hydrogen chloride (hydrochloric acid) and oxygen.
(h) For the reaction 2H2S + SO2 → 3S + 2H2O
Assigning oxidation numbers to sulphur: it is −2 in H2S, +4 in SO2, and 0 in the free sulphur produced.
- Species undergoing oxidation: H2S - the sulphur in it rises from −2 to 0 (loses electrons).
- Oxidizing agent: SO2 - the sulphur in it falls from +4 to 0 (it is reduced, so it oxidizes the H2S).
(i) Carbon-12 scale
It is the scale of relative atomic masses on which one atom of the carbon-12 isotope is assigned a mass of exactly 12.000 units, and the mass of every other atom is measured relative to one-twelfth of the mass of a carbon-12 atom.
(j) Two properties of a chemical system in equilibrium
- It is dynamic: the forward and backward reactions continue at equal rates.
- Its macroscopic properties (concentrations, colour, pressure) remain constant, and it is attained only in a closed system.
Ajụjụ 5 Ripọtì
(a) (i) What is meant by hardness of water?
(ii) Describe briefly how you would determine what proportion of hardness in a given sample of water is due to permanent hardness.
(iii) Give two reasons why hardness of water is an undesirable property.
(b) State the:
(i) reagents;
(ii) conditions for the laboratory preparation of trioxonitrate (V) acid.
(iii) How does concentrated trioxonitrate (V) acid reacts with:
(I) sulphur;
(II) aluminium.
(c) Name one amphoteric oxide.
(a)(i) Hardness of water is the property of water that prevents it from readily forming lather (foam) with soap, caused by dissolved calcium and magnesium salts.
(ii) Boil a measured sample of the water to remove temporary hardness, then filter. Determine (by soap-lather titration) the soap needed to give a lasting lather with the unboiled sample (total hardness) and with the boiled sample (permanent hardness). The permanent hardness is the value obtained with the boiled sample; the difference gives the temporary hardness.
(iii) Any two: it wastes soap and forms scum; it deposits scale (fur) in kettles, boilers and pipes, which wastes fuel and can block pipes.
(b)(i) Reagents: potassium (or sodium) trioxonitrate(V), KNO3, and concentrated tetraoxosulphate(VI) acid.
(ii) Conditions: gentle heating, using all-glass apparatus (the acid attacks cork and rubber).
(iii) Concentrated HNO3 with:
- Sulphur: it oxidizes sulphur to tetraoxosulphate(VI) acid. \(S + 6HNO_3 \rightarrow H_2SO_4 + 6NO_2 + 2H_2O\)
- Aluminium: no reaction; aluminium is rendered passive because a protective oxide layer forms on its surface.
(c) One amphoteric oxide: aluminium oxide, Al2O3 (also ZnO or PbO).
Akọwa Nkọwa
(a)(i) Hardness of water is the property of water that prevents it from readily forming lather (foam) with soap, caused by dissolved calcium and magnesium salts.
(ii) Boil a measured sample of the water to remove temporary hardness, then filter. Determine (by soap-lather titration) the soap needed to give a lasting lather with the unboiled sample (total hardness) and with the boiled sample (permanent hardness). The permanent hardness is the value obtained with the boiled sample; the difference gives the temporary hardness.
(iii) Any two: it wastes soap and forms scum; it deposits scale (fur) in kettles, boilers and pipes, which wastes fuel and can block pipes.
(b)(i) Reagents: potassium (or sodium) trioxonitrate(V), KNO3, and concentrated tetraoxosulphate(VI) acid.
(ii) Conditions: gentle heating, using all-glass apparatus (the acid attacks cork and rubber).
(iii) Concentrated HNO3 with:
- Sulphur: it oxidizes sulphur to tetraoxosulphate(VI) acid. \(S + 6HNO_3 \rightarrow H_2SO_4 + 6NO_2 + 2H_2O\)
- Aluminium: no reaction; aluminium is rendered passive because a protective oxide layer forms on its surface.
(c) One amphoteric oxide: aluminium oxide, Al2O3 (also ZnO or PbO).
Ajụjụ 6 Ripọtì
(a) (i) Write an equation for the reaction by which sulphur dioxide in solution could be converted to tetraoxosulphate (VI) acid.
(ii) State one test to confirm the conversion of sulphur dioxide to tetraoxosulphate (VI) acid.
(iii) State the reaction of concentrated tetraoxosulphate (VI) acid with:
(I) oxalic acid;
(II) copper.
(iv) What property of concentrated tetraoxosulphate (VI) acid does each of the reactions stated in (a)(iii) illustrate?
(b) (i) Explain briefly why water is referred to as a universal solvent.
(ii) Give one chemical test for water.
(c) (i) What is the major component of synthetic gas?
(ii) Give one reason why synthetic gas is not a major source of air pollution.
(d) Name one product of destructive distillation of coal that is:
(i) solid;
(ii) liquid;
(iii) gas.
(e) (i) Write a balanced chemical equation for the complete combustion of carbon.
(ii) State one:
(I) physical;
(II) chemical property of the products in (e)(i).
(a)(i) \(SO_2 + Cl_2 + 2H_2O \rightarrow H_2SO_4 + 2HCl\) (sulphur dioxide is oxidized to tetraoxosulphate(VI) acid).
(ii) Add barium chloride solution followed by dilute HCl: a white precipitate of BaSO4 (insoluble in the acid) confirms the tetraoxosulphate(VI) ion.
(iii) Concentrated H2SO4 with:
- Oxalic acid: it dehydrates the acid. \((COOH)_2 \rightarrow CO + CO_2 + H_2O\)
- Copper (hot): it oxidizes copper. \(Cu + 2H_2SO_4 \rightarrow CuSO_4 + SO_2 + 2H_2O\)
(iv) With oxalic acid it acts as a dehydrating agent; with copper it acts as an oxidizing agent.
(b)(i) Water is called a universal solvent because it dissolves a greater number of substances than any other solvent, owing to its polar nature.
(ii) Add the liquid to white anhydrous copper(II) tetraoxosulphate(VI): it turns blue if water is present (or anhydrous cobalt(II) chloride turns from blue to pink).
(c)(i) The major components of synthesis gas are carbon(II) oxide and hydrogen.
(ii) It burns completely and cleanly to give carbon(IV) oxide and water, producing no soot or sulphurous fumes.
(d) From destructive distillation of coal: (i) solid: coke; (ii) liquid: coal tar (or ammoniacal liquor); (iii) gas: coal gas.
(e)(i) \(C + O_2 \rightarrow CO_2\)
(ii) Carbon(IV) oxide: (I) physical: it is a colourless gas, denser than air; (II) chemical: it turns lime water milky and does not support combustion.
Akọwa Nkọwa
(a)(i) \(SO_2 + Cl_2 + 2H_2O \rightarrow H_2SO_4 + 2HCl\) (sulphur dioxide is oxidized to tetraoxosulphate(VI) acid).
(ii) Add barium chloride solution followed by dilute HCl: a white precipitate of BaSO4 (insoluble in the acid) confirms the tetraoxosulphate(VI) ion.
(iii) Concentrated H2SO4 with:
- Oxalic acid: it dehydrates the acid. \((COOH)_2 \rightarrow CO + CO_2 + H_2O\)
- Copper (hot): it oxidizes copper. \(Cu + 2H_2SO_4 \rightarrow CuSO_4 + SO_2 + 2H_2O\)
(iv) With oxalic acid it acts as a dehydrating agent; with copper it acts as an oxidizing agent.
(b)(i) Water is called a universal solvent because it dissolves a greater number of substances than any other solvent, owing to its polar nature.
(ii) Add the liquid to white anhydrous copper(II) tetraoxosulphate(VI): it turns blue if water is present (or anhydrous cobalt(II) chloride turns from blue to pink).
(c)(i) The major components of synthesis gas are carbon(II) oxide and hydrogen.
(ii) It burns completely and cleanly to give carbon(IV) oxide and water, producing no soot or sulphurous fumes.
(d) From destructive distillation of coal: (i) solid: coke; (ii) liquid: coal tar (or ammoniacal liquor); (iii) gas: coal gas.
(e)(i) \(C + O_2 \rightarrow CO_2\)
(ii) Carbon(IV) oxide: (I) physical: it is a colourless gas, denser than air; (II) chemical: it turns lime water milky and does not support combustion.
Ajụjụ 7 Ripọtì
(a) (i) State the important points put forward in Dalton’s atomic theory.
(ii) How does the theory explain the law of multiple proportion?
(b) (i) Name three commercially useful products that may be obtained by chemical transformation of vegetable oils.
(ii) With the aid of equations, describe the chemical reaction involved in the transformations stated in (b)(i).
(c) Certain properties of Beryllium (Be) and its compounds differ from those of Magnesium (Mg) and its compounds, but rather resembles those of Aluminium and its compounds. Explain briefly why this is so.
(d) Write balanced equation for the reaction between iodine and aqueous sodium trioxothiosulphate(VI).
(a)(i) Dalton's atomic theory: matter is made of tiny indivisible particles called atoms; atoms of the same element are identical in mass and properties, and differ from those of other elements; atoms combine in simple whole-number ratios to form compounds; atoms are neither created nor destroyed in a chemical reaction (which is only a rearrangement of atoms).
(ii) Because whole atoms combine in simple whole-number ratios, when two elements form more than one compound the masses of one element that combine with a fixed mass of the other must be in a simple whole-number ratio, which is the law of multiple proportions.
(b)(i) Three products: margarine (by hydrogenation), soap (by saponification), and glycerol (by hydrolysis/saponification).
(ii)
- Hydrogenation (to margarine): unsaturated oil \(+ H_2 \xrightarrow{Ni} \) hardened (saturated) fat.
- Saponification (to soap): \((C_{17}H_{35}COO)_3C_3H_5 + 3NaOH \rightarrow 3C_{17}H_{35}COONa + C_3H_5(OH)_3\)
- Hydrolysis (to glycerol): \((C_{17}H_{35}COO)_3C_3H_5 + 3H_2O \rightarrow 3C_{17}H_{35}COOH + C_3H_5(OH)_3\)
(c) This is a diagonal relationship: beryllium and aluminium lie diagonally in the periodic table and have similar charge-to-radius ratios (similar polarizing power and electronegativity), so their properties resemble each other rather than those of magnesium.
(d) \(I_2 + 2Na_2S_2O_3 \rightarrow 2NaI + Na_2S_4O_6\)
Akọwa Nkọwa
(a)(i) Dalton's atomic theory: matter is made of tiny indivisible particles called atoms; atoms of the same element are identical in mass and properties, and differ from those of other elements; atoms combine in simple whole-number ratios to form compounds; atoms are neither created nor destroyed in a chemical reaction (which is only a rearrangement of atoms).
(ii) Because whole atoms combine in simple whole-number ratios, when two elements form more than one compound the masses of one element that combine with a fixed mass of the other must be in a simple whole-number ratio, which is the law of multiple proportions.
(b)(i) Three products: margarine (by hydrogenation), soap (by saponification), and glycerol (by hydrolysis/saponification).
(ii)
- Hydrogenation (to margarine): unsaturated oil \(+ H_2 \xrightarrow{Ni} \) hardened (saturated) fat.
- Saponification (to soap): \((C_{17}H_{35}COO)_3C_3H_5 + 3NaOH \rightarrow 3C_{17}H_{35}COONa + C_3H_5(OH)_3\)
- Hydrolysis (to glycerol): \((C_{17}H_{35}COO)_3C_3H_5 + 3H_2O \rightarrow 3C_{17}H_{35}COOH + C_3H_5(OH)_3\)
(c) This is a diagonal relationship: beryllium and aluminium lie diagonally in the periodic table and have similar charge-to-radius ratios (similar polarizing power and electronegativity), so their properties resemble each other rather than those of magnesium.
(d) \(I_2 + 2Na_2S_2O_3 \rightarrow 2NaI + Na_2S_4O_6\)
Ajụjụ 8 Ripọtì
(a) The following reaction scheme is an illustration of the contact process. Study the scheme and answer the questions that follow.
(i) Name X and Y
(ii) Write a balanced chemical equation for each of the processes I, II, III and IV
(iii) Name the catalyst used in process II
(iv) Using Le Chatelier's principle, explain briefly why increasing the temperature would not favour the reaction in II
(v) State two uses of \(SO_2\)
(b) Consider the following equation: \(2H_{2(g)} + O_{2(g)} \to 2H_2O_{(g)}\)
Calculate the volume of unused oxygen gas when \(40\ \text{cm}^3\) of hydrogen gas is sparked with \(30\text{cm}^3\) of oxygen gas
(c) Calcium carbonate of mass 1.0 g was heated until there was no further change.
- Write an equation for the reaction which took place.
- Calculate the mass of the residue.
- Calculate the volume of the gas evolved at s.t.p.
- What would be the volume of the gas measured at 15 and 760 mm Hg? [C= 12.0, O = 16.0, Ca = 40.0, molar volume of a gas at s.t.p. = 22.4 dm\(^3\)]
(a) The Contact process. The scheme reads: X(gas) + Y(solid) \(\xrightarrow{I}\) Sulphur(IV) oxide; Oxygen + Sulphur(IV) oxide \(\xrightarrow{II}\) Sulphur(VI) oxide \(\xrightarrow{III}\) Oleum \(\xrightarrow{IV}\) Concentrated H2SO4.
(i) X and Y. The first step burns a solid in a gas to give SO2, so X is oxygen (air) and Y is sulphur.
(ii) Balanced equations.
Process I: \[ S_{(s)} + O_{2(g)} \rightarrow SO_{2(g)} \]
Process II: \[ 2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)} \]
Process III (SO3 absorbed in concentrated H2SO4 to form oleum): \[ SO_{3(g)} + H_2SO_{4(l)} \rightarrow H_2S_2O_{7(l)} \]
Process IV (oleum diluted with water): \[ H_2S_2O_{7(l)} + H_2O_{(l)} \rightarrow 2H_2SO_{4(l)} \]
(iii) Catalyst in process II: vanadium(V) oxide, V2O5 (platinum can also be used).
(iv) Le Chatelier's principle. Process II is exothermic in the forward direction. By Le Chatelier's principle, raising the temperature shifts the equilibrium position in the endothermic (backward) direction so as to absorb the added heat. This reduces the yield of SO3, so a high temperature does not favour the forward reaction; a moderate temperature (about 450 C) is used instead.
(v) Two uses of SO2:
- As a bleaching agent (for wood pulp, straw, silk).
- As a food preservative and fumigant/disinfectant (it also serves as the starting gas for making H2SO4).
(b) Volume of unused oxygen. From \(2H_{2(g)} + O_{2(g)} \rightarrow 2H_2O_{(g)}\), H2 reacts with O2 in the ratio 2 : 1 by volume.
\[ \text{O}_2\ \text{needed} = \frac{1}{2} \times 40 = 20\ \text{cm}^3 \]
\[ \text{Unused O}_2 = 30 - 20 = \mathbf{10\ cm^3} \]
(c) Heating calcium trioxocarbonate(IV).
\[ CaCO_{3(s)} \rightarrow CaO_{(s)} + CO_{2(g)} \]
Moles of CaCO3: molar mass \(= 40.0 + 12.0 + (3\times16.0) = 100\ g\,mol^{-1}\).
\[ n = \frac{1.0}{100} = 0.01\ \text{mol} \]
Mass of residue (CaO): molar mass of CaO \(= 40.0 + 16.0 = 56\ g\,mol^{-1}\).
\[ m_{CaO} = 0.01 \times 56 = \mathbf{0.56\ g} \]
Volume of CO2 at s.t.p.: \(n_{CO_2} = 0.01\ mol\).
\[ V = 0.01 \times 22.4 = 0.224\ dm^3 = \mathbf{224\ cm^3} \]
Volume at 15 C and 760 mmHg: pressure is unchanged (760 mmHg = s.t.p. pressure), so use \(\dfrac{V_1}{T_1} = \dfrac{V_2}{T_2}\) with \(T_1 = 273\ K,\ T_2 = 273 + 15 = 288\ K\).
\[ V_2 = 0.224 \times \frac{288}{273} = \mathbf{0.236\ dm^3\ (\approx 236\ cm^3)} \]
Akọwa Nkọwa
(a) The Contact process. The scheme reads: X(gas) + Y(solid) \(\xrightarrow{I}\) Sulphur(IV) oxide; Oxygen + Sulphur(IV) oxide \(\xrightarrow{II}\) Sulphur(VI) oxide \(\xrightarrow{III}\) Oleum \(\xrightarrow{IV}\) Concentrated H2SO4.
(i) X and Y. The first step burns a solid in a gas to give SO2, so X is oxygen (air) and Y is sulphur.
(ii) Balanced equations.
Process I: \[ S_{(s)} + O_{2(g)} \rightarrow SO_{2(g)} \]
Process II: \[ 2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)} \]
Process III (SO3 absorbed in concentrated H2SO4 to form oleum): \[ SO_{3(g)} + H_2SO_{4(l)} \rightarrow H_2S_2O_{7(l)} \]
Process IV (oleum diluted with water): \[ H_2S_2O_{7(l)} + H_2O_{(l)} \rightarrow 2H_2SO_{4(l)} \]
(iii) Catalyst in process II: vanadium(V) oxide, V2O5 (platinum can also be used).
(iv) Le Chatelier's principle. Process II is exothermic in the forward direction. By Le Chatelier's principle, raising the temperature shifts the equilibrium position in the endothermic (backward) direction so as to absorb the added heat. This reduces the yield of SO3, so a high temperature does not favour the forward reaction; a moderate temperature (about 450 C) is used instead.
(v) Two uses of SO2:
- As a bleaching agent (for wood pulp, straw, silk).
- As a food preservative and fumigant/disinfectant (it also serves as the starting gas for making H2SO4).
(b) Volume of unused oxygen. From \(2H_{2(g)} + O_{2(g)} \rightarrow 2H_2O_{(g)}\), H2 reacts with O2 in the ratio 2 : 1 by volume.
\[ \text{O}_2\ \text{needed} = \frac{1}{2} \times 40 = 20\ \text{cm}^3 \]
\[ \text{Unused O}_2 = 30 - 20 = \mathbf{10\ cm^3} \]
(c) Heating calcium trioxocarbonate(IV).
\[ CaCO_{3(s)} \rightarrow CaO_{(s)} + CO_{2(g)} \]
Moles of CaCO3: molar mass \(= 40.0 + 12.0 + (3\times16.0) = 100\ g\,mol^{-1}\).
\[ n = \frac{1.0}{100} = 0.01\ \text{mol} \]
Mass of residue (CaO): molar mass of CaO \(= 40.0 + 16.0 = 56\ g\,mol^{-1}\).
\[ m_{CaO} = 0.01 \times 56 = \mathbf{0.56\ g} \]
Volume of CO2 at s.t.p.: \(n_{CO_2} = 0.01\ mol\).
\[ V = 0.01 \times 22.4 = 0.224\ dm^3 = \mathbf{224\ cm^3} \]
Volume at 15 C and 760 mmHg: pressure is unchanged (760 mmHg = s.t.p. pressure), so use \(\dfrac{V_1}{T_1} = \dfrac{V_2}{T_2}\) with \(T_1 = 273\ K,\ T_2 = 273 + 15 = 288\ K\).
\[ V_2 = 0.224 \times \frac{288}{273} = \mathbf{0.236\ dm^3\ (\approx 236\ cm^3)} \]
Ajụjụ 9 Ripọtì
(a)(i) Draw and label a diagram to illustrate the preparation and collection of dry chlorine gas in the laboratory.
(ii) State two uses of chlorine.
(b) Describe the preparation of hydrogen from water gas.
(i) Name the chief ore of aluminium.
(ii) Why is the ore purified?
(iii) Name the electrode used in the electrolysis.
(iv) Give one reason why cryolite, \( \mathrm{NaAlF_6} \), is added to the electrolyte.
(c) Name three products obtained directly from the destructive distillation of coal.
(a)(i) Laboratory preparation and collection of dry chlorine gas
Chlorine is prepared by heating manganese(IV) oxide with concentrated hydrochloric acid:
\[MnO_2(s)+4HCl(aq)\rightarrow MnCl_2(aq)+Cl_2(g)+2H_2O(l)\]
The gas is passed through saturated sodium chloride solution to remove hydrogen chloride, then through concentrated sulphuric acid to dry it. It is collected by downward delivery, that is, by upward displacement of air, because chlorine is denser than air and dissolves in water.
(ii) Uses of chlorine
- It is used to disinfect and sterilize drinking water.
- It is used as a bleaching agent for paper pulp and textiles.
(b) Preparation of hydrogen from water gas
Water gas, a mixture of carbon(II) oxide and hydrogen, is mixed with steam and passed over heated iron(III) oxide or chromium(III) oxide catalyst at about \(450^\circ\mathrm{C}\). The carbon(II) oxide is converted to carbon(IV) oxide, producing more hydrogen:
\[CO(g)+H_2O(g)\rightarrow CO_2(g)+H_2(g)\]
The carbon(IV) oxide is removed by absorption in water under pressure or in aqueous sodium hydroxide, potassium hydroxide, or calcium hydroxide. The gas remaining is hydrogen.
(c)(i) The chief ore of aluminium is bauxite, \(Al_2O_3\cdot 2H_2O\).
(ii) The ore is purified to obtain alumina and to remove impurities, such as iron(III) oxide and silica, which would contaminate the aluminium and may poison the electrodes.
(iii) The electrodes used are carbon (graphite) electrodes.
(iv) Cryolite, \(Na_3AlF_6\), is added to lower the melting point of alumina, thereby reducing the energy required for electrolysis. It also increases the electrical conductivity of the electrolyte.
(d) Three products obtained directly from the destructive distillation of coal are:
- coke;
- coal tar;
- coal gas.
Ammoniacal liquor is also obtained.
Akọwa Nkọwa
(a)(i) Laboratory preparation and collection of dry chlorine gas
Chlorine is prepared by heating manganese(IV) oxide with concentrated hydrochloric acid:
\[MnO_2(s)+4HCl(aq)\rightarrow MnCl_2(aq)+Cl_2(g)+2H_2O(l)\]
The gas is passed through saturated sodium chloride solution to remove hydrogen chloride, then through concentrated sulphuric acid to dry it. It is collected by downward delivery, that is, by upward displacement of air, because chlorine is denser than air and dissolves in water.
(ii) Uses of chlorine
- It is used to disinfect and sterilize drinking water.
- It is used as a bleaching agent for paper pulp and textiles.
(b) Preparation of hydrogen from water gas
Water gas, a mixture of carbon(II) oxide and hydrogen, is mixed with steam and passed over heated iron(III) oxide or chromium(III) oxide catalyst at about \(450^\circ\mathrm{C}\). The carbon(II) oxide is converted to carbon(IV) oxide, producing more hydrogen:
\[CO(g)+H_2O(g)\rightarrow CO_2(g)+H_2(g)\]
The carbon(IV) oxide is removed by absorption in water under pressure or in aqueous sodium hydroxide, potassium hydroxide, or calcium hydroxide. The gas remaining is hydrogen.
(c)(i) The chief ore of aluminium is bauxite, \(Al_2O_3\cdot 2H_2O\).
(ii) The ore is purified to obtain alumina and to remove impurities, such as iron(III) oxide and silica, which would contaminate the aluminium and may poison the electrodes.
(iii) The electrodes used are carbon (graphite) electrodes.
(iv) Cryolite, \(Na_3AlF_6\), is added to lower the melting point of alumina, thereby reducing the energy required for electrolysis. It also increases the electrical conductivity of the electrolyte.
(d) Three products obtained directly from the destructive distillation of coal are:
- coke;
- coal tar;
- coal gas.
Ammoniacal liquor is also obtained.
Ajụjụ 10 Ripọtì
(a) (i) Name three different methods for preparing salts.
(ii) Give one example of a balanced equation for each of the methods named in (a)(i).
(iii) State two uses of sodium trioxocarbonate (IV).
(b) If you were given some impure copper, describe how you would obtain a specimen of the pure metal by electrolysis.
(c) Given that sodium chloride has a solubility of 36.3 at 30 and 39.0 at 100 and that of silver nitrate is 297.0 at 30 and 952.0 at 100.
(i) Calculate the percentage of each substance in the saturated solution at 100 that is deposited on cooling to 30
(ii) Deduce which of the two salts can be purified more efficiently by crystallization.
(a)(i) Three methods: neutralization (acid + alkali/base); action of an acid on a metal; precipitation (double decomposition).
(ii) One equation for each:
- Neutralization: \(NaOH + HCl \rightarrow NaCl + H_2O\)
- Acid on metal: \(Zn + H_2SO_4 \rightarrow ZnSO_4 + H_2\)
- Precipitation: \(AgNO_3 + NaCl \rightarrow AgCl + NaNO_3\)
(iii) Two uses of sodium trioxocarbonate(IV), Na2CO3: manufacture of glass; softening of hard water (also soap/detergent and paper manufacture).
(b) Purifying copper by electrolysis: use the impure copper as the anode and a thin strip of pure copper as the cathode, dipped in acidified copper(II) tetraoxosulphate(VI) solution as electrolyte. On passing current, the anode dissolves (\(Cu \rightarrow Cu^{2+} + 2e^-\)) and pure copper is deposited on the cathode (\(Cu^{2+} + 2e^- \rightarrow Cu\)). Impurities collect below the anode as anode sludge.
(c)(i) Take the solubility (g per 100 g water) as the mass dissolved at each temperature.
Sodium chloride: deposited \(= 39.0 - 36.3 = 2.7\) g.
Percentage of the dissolved salt deposited \(= \dfrac{2.7}{39.0} \times 100 = \mathbf{6.9\%}\).
Silver trioxonitrate(V): deposited \(= 952.0 - 297.0 = 655.0\) g.
Percentage of the dissolved salt deposited \(= \dfrac{655.0}{952.0} \times 100 = \mathbf{68.8\%}\).
(ii) Silver trioxonitrate(V) can be purified more efficiently by crystallization, because a far larger fraction of it (about 68.8%) crystallizes out on cooling than for sodium chloride (about 6.9%).
Akọwa Nkọwa
(a)(i) Three methods: neutralization (acid + alkali/base); action of an acid on a metal; precipitation (double decomposition).
(ii) One equation for each:
- Neutralization: \(NaOH + HCl \rightarrow NaCl + H_2O\)
- Acid on metal: \(Zn + H_2SO_4 \rightarrow ZnSO_4 + H_2\)
- Precipitation: \(AgNO_3 + NaCl \rightarrow AgCl + NaNO_3\)
(iii) Two uses of sodium trioxocarbonate(IV), Na2CO3: manufacture of glass; softening of hard water (also soap/detergent and paper manufacture).
(b) Purifying copper by electrolysis: use the impure copper as the anode and a thin strip of pure copper as the cathode, dipped in acidified copper(II) tetraoxosulphate(VI) solution as electrolyte. On passing current, the anode dissolves (\(Cu \rightarrow Cu^{2+} + 2e^-\)) and pure copper is deposited on the cathode (\(Cu^{2+} + 2e^- \rightarrow Cu\)). Impurities collect below the anode as anode sludge.
(c)(i) Take the solubility (g per 100 g water) as the mass dissolved at each temperature.
Sodium chloride: deposited \(= 39.0 - 36.3 = 2.7\) g.
Percentage of the dissolved salt deposited \(= \dfrac{2.7}{39.0} \times 100 = \mathbf{6.9\%}\).
Silver trioxonitrate(V): deposited \(= 952.0 - 297.0 = 655.0\) g.
Percentage of the dissolved salt deposited \(= \dfrac{655.0}{952.0} \times 100 = \mathbf{68.8\%}\).
(ii) Silver trioxonitrate(V) can be purified more efficiently by crystallization, because a far larger fraction of it (about 68.8%) crystallizes out on cooling than for sodium chloride (about 6.9%).
