4ch1-1c-que-20231115

Question 1

A substance can exist in three states of matter, solid, liquid or gas. When a liquid evaporates at room temperature it changes into a gas.

(a) Complete the diagram to show the arrangement of another four particles in a gas.

(b) Explain why heating a liquid causes it to evaporate more quickly.

(c) When the temperature decreases, water in the gas state changes to a liquid.
(i) Give the name of this change of state.
(ii) Write an equation, including state symbols, to show the change of state of water from a gas to a liquid.

(d) Describe the arrangement and movement of particles in a solid.

Most-appropriate topic codes (Edexcel IGCSE Chemistry 4CH1):

• 1(a): States of matter — All parts
• Specification Points: 1.1 (understand the three states of matter…), 1.2 (understand the interconversions…).

▶️ Answer/Explanation

Solution

(a) Four more particles (randomly spaced and) far apart.

Marking Note: Particles must be far apart and not touching. Accept more than four.

(b) An explanation linking:
• M1: (Mean kinetic) energy of particles increases / particles move faster.
• M2: More particles have enough energy to escape / overcome the intermolecular forces.

Marking Note: Allow “molecules” for “particles”. Do not accept “breaking bonds”.

(c)(i) Condensation / condensing.

(c)(ii) $ \text{H}_2\text{O(g)} \rightarrow \text{H}_2\text{O(l)} $
Marking Note: Accept formulas with correct state symbols, e.g., water vapour(g) → water(l).

(d) Description including:
• M1: Regular arrangement of particles / particles closely packed.
• M2: (Particles) vibrate around a fixed position / vibrate only.

Marking Note: Do not accept “particles move freely”.

Question 2

(a) Table 1 shows some relative masses and charges of subatomic particles. Complete Table 1 by giving the missing information.

	Electron	Proton	Neutron
Relative mass	0.0005
Relative charge			0

(b) Table 2 gives the number of protons, neutrons, and electrons in atoms and ions of some elements. The letters are not the symbols of the elements.

Atom or ion	Protons	Neutrons	Electrons
P	3	4	2
Q	5	5	5
R	5	6	5
S	7	7	7
T	8	8	8
U	8	8	10

(i) What is the atomic number of P in Table 2?
A) 2
B) 3
C) 4
D) 7

(ii) What is the mass number of U in Table 2?
A) 8
B) 16
C) 18
D) 26

(iii) Give the letter in Table 2 that represents an element in Group 5 of the Periodic Table.

(c) Q and R represent isotopes of the same element.

(i) Explain, in terms of subatomic particles, why Q and R are isotopes.
(ii) A sample containing the isotopes Q and R has this percentage composition by mass:

Q = 20.6% R = 79.4%

Calculate the relative atomic mass ($A_r$) of this sample of the element. Give your answer to one decimal place.

Most-appropriate topic codes (Edexcel IGCSE Chemistry 4CH1):

• 1(c): Atomic structure — parts (a), (b)(i), (b)(ii), (c)(i), (c)(ii) (specifically 1.15, 1.16, 1.17)
• 1(d): The Periodic Table — part (b)(iii) (specifically 1.18, 1.22, 1.23)

▶️ Answer/Explanation

Solution

(a) Completed Table:

	Electron	Proton	Neutron
Relative mass	0.0005	1	1
Relative charge	–1	+1	0

(b)(i) B (3)
• A is incorrect as 2 is not the atomic number of P.
• C is incorrect as 4 is not the atomic number of P.
• D is incorrect as 7 is the mass number of P.

(b)(ii) B (16)
• A is incorrect as 8 is the atomic number of U.
• C is incorrect as 18 is not the mass number of U.
• D is incorrect as 26 is not the mass number of U.
• Mass number = protons + neutrons = 8 + 8 = 16.

(b)(iii) S
• S has 7 protons, so its atomic number is 7. Group 5 elements have 5 electrons in their outer shell, which corresponds to an electronic configuration ending with 5 electrons in the highest energy level. Nitrogen (atomic number 7) is in Group 5.

(c)(i) An explanation that links the following two points:
• (Q and R have) the same number of protons / both have 5 protons. (1)
• (but) different numbers of neutrons / (Q has) 5 neutrons and (R has) 6 neutrons / R has an extra neutron. (1)

(c)(ii) Calculation of relative atomic mass:
Method 1:
$ \text{Average mass} = \frac{(20.6 \times 10) + (79.4 \times 11)}{100} $
$ = \frac{206 + 873.4}{100} = \frac{1079.4}{100} = 10.794 $

Method 2:
$ \text{Average mass} = \left(\frac{20.6}{100} \times 10\right) + \left(\frac{79.4}{100} \times 11\right) $
$ = (0.206 \times 10) + (0.794 \times 11) = 2.06 + 8.734 = 10.794 $

Rounded to one decimal place: $ \mathbf{10.8} $
Note: Mass number of Q = 5 protons + 5 neutrons = 10. Mass number of R = 5 protons + 6 neutrons = 11.

Question 3

Crude oil is an important source of organic compounds.

(a) The diagram shows how crude oil can be separated into useful mixtures of hydrocarbons.

(i) Give the name of this method of separation.
(ii) State what happens to the crude oil when it is in X.
(iii) Give the letter of the mixture that is most likely to contain a hydrocarbon with six carbon atoms.
(iv) Give the name of mixture D.
(v) Give a use for mixture B.

(b) Explain, in terms of intermolecular forces, why a hydrocarbon in mixture B has a higher boiling point than a hydrocarbon in mixture D.

(c) Catalytic cracking can be used to break down long-chain hydrocarbons to produce shorter-chain alkanes and alkenes.

(i) Give the name of a catalyst used in catalytic cracking.

(ii) Complete the equation to show two different alkenes that could be produced in this cracking reaction.

$$ \text{C}_{15}\text{H}_{32} \rightarrow \text{C}_{8}\text{H}_{18} + \text{……} + \text{……} $$

(iii) Give one important use for short-chain alkenes.

Most-appropriate topic codes (Edexcel IGCSE Chemistry 4CH1):

• 4(b): Crude oil — parts (a)(i)-(v), (c)(i)-(ii)
• 1(g): Covalent bonding & Intermolecular forces — part (b)
• 4(d): Alkenes — part (c)(iii)

▶️ Answer/Explanation

Solution

(a)(i) fractional distillation
ALLOW fractionating

(a)(ii) (crude oil/it is) heated / vapourised/ boiled
Ignore evaporated

(a)(iii) E
ALLOW gasoline/petrol

(a)(iv) kerosene
ALLOW paraffin

(a)(v) (Fuel) for ships
ALLOW any acceptable use of fuel oil e.g. home heating, industrial heating, electricity generation, power station, furnaces for metal smelting, feedstock for plastics/fertilisers

(b) An explanation that links the following three points:
• M1: B has longer chain/molecules ORA
ALLOW B has larger/bigger/longer chain/molecule/hydrocarbon OR molecule/hydrocarbon with greater mass
• M2: B has stronger intermolecular forces/bonds
ALLOW more intermolecular forces/bonds. Forces/bonds between molecules ORA. REJECT IMF between atoms.
• M3: more energy is needed to overcome the (intermolecular) forces/intermolecular bonds separate the molecules ORA
No M2 or M3 if any reference to breaking of covalent bonds.

(c)(i) silica / alumina (catalyst)
ACCEPT $\text{SiO}_2$ / $\text{Al}_2\text{O}_3$ / silicon dioxide / aluminium oxide / aluminosilicates / zeolites

(c)(ii) Any one of the following two pairs:
• M1: $\text{C}_2\text{H}_4$ and M2: $\text{C}_5\text{H}_{10}$ OR
• M1: $\text{C}_3\text{H}_6$ and M2: $\text{C}_4\text{H}_8$
If the equation does not balance allow 1 mark for a correct formula of an alkene.

(c)(iii) (to make) polymers / polymerisation
ACCEPT the name of a correct addition polymer e.g. polyethene, polypropene etc.
ACCEPT to make alcohol(s). Reject fuels.

Total marks: 12

Question 4

This question is about gases.

(a) The table gives information about five gases.

Name of gas	argon	carbon dioxide	hydrogen	oxygen	nitrogen
Formula of gas	Ar	CO₂	H₂	O₂	N₂
M_r of gas	40		2	32	28

Use information from the table to answer these questions. Each gas may be used once, more than once or not at all.

(i) Give the name of the gas that is about 79% of the atmosphere by volume.
(ii) Give the name of the gas that is a compound.
(iii) Give the name of the least reactive gas.
(iv) Give the name of the gas that is not normally found in the atmosphere.
(v) Give the name of the gas that affects global warming.
(vi) Calculate the M_r for carbon dioxide.
(vii) Give a reason why it is not possible to give information for air in the table.

(b) When copper(II) carbonate is heated, the products are copper(II) oxide and carbon dioxide.

(i) Give the name for this type of reaction.
(ii) Give the colour change that occurs during this reaction.
(iii) Give a chemical equation for this reaction.

Most-appropriate topic codes (Edexcel IGCSE Chemistry 4CH1):

• 2(c): Gases in the atmosphere — parts (a)(i), (a)(iii), (a)(iv), (a)(v), (a)(vii)
• 1(b): Elements, compounds and mixtures — part (a)(ii)
• 1(e): Chemical formulae, equations and calculations — part (a)(vi)
• 2(c): Thermal decomposition of metal carbonates — part (b)(i), (b)(ii), (b)(iii)

▶️ Answer/Explanation

Solution

(a)(i) nitrogen
ALLOW N₂

(a)(ii) carbon dioxide
ALLOW CO₂

(a)(iii) argon
ALLOW Ar

(a)(iv) hydrogen
ALLOW H₂

(a)(v) carbon dioxide
ALLOW CO₂

(a)(vi) $ M_r = (12 + 2 \times 16) = 44 $

(a)(vii) air is a mixture (of gases) / does not have a formula / does not have an $ M_r $ / OWTTE

(b)(i) (thermal) decomposition

(b)(ii) green (1) to black (1)

(b)(iii) $ CuCO_3 \rightarrow CuO + CO_2 $
ALLOW multiples and fractions.
IGNORE state symbols even if incorrect.

Total = 11 marks

Question 5

This question is about alkanes and alkenes.

(a) The alkane $ C_4H_{10} $ exists as two isomers.

(i) State what is meant by the term isomers.

(ii) Draw the displayed formulae for the two isomers of $ C_4H_{10} $.

Isomer 1	Isomer 2

(b) Ethane ($ C_2H_6 $) can react with bromine.

(i) State the condition needed for ethane to react with bromine.

(ii) Complete the equation for this reaction.

$C_2H_6 + Br_2 \rightarrow $ ______ + ______

(iii) Give the name for this type of reaction.

(c) Explain why ethane is described as a saturated compound.

(d) State what you would observe when ethane and ethene are added separately to two samples of bromine water.

(e) Explain why straight-chain alkenes always have the same empirical formula, but straight-chain alkanes have different empirical formulae.
Refer to the molecular formulae of the alkanes $ C_2H_6 $ and $ C_4H_{10} $ in your answer.

(f) An organic compound has this percentage composition by mass.

$ C = 19.2\% \quad H = 4.0\% \quad O = 12.8\% \quad Br = 64.0\% $

Calculate the empirical formula of this compound.

Most-appropriate topic codes (Edexcel IGCSE Chemistry):

• 4(a): Introduction (to Organic Chemistry) — parts (a)(i), (a)(ii), (b)(iii)
• 4(c): Alkanes — parts (a)(ii), (b)(i), (b)(ii), (c)
• 4(d): Alkenes — parts (d), (e)
• 1(e): Chemical formulae, equations and calculations — part (f)
• 1(g): Covalent bonding — part (a)(ii)

▶️ Answer/Explanation

Solution

(a)(i)
M1: (Compounds with) the same molecular formula.
M2: But different structural/displayed formulae / different arrangement of atoms.

(a)(ii)
M1 (Isomer 1 – Butane): CH₃–CH₂–CH₂–CH₃ (displayed formula showing all atoms and bonds).

M2 (Isomer 2 – Methylpropane): CH₃–CH(CH₃)–CH₃ (displayed formula showing all atoms and bonds).

(b)(i) Ultraviolet (UV) radiation / UV light.

(b)(ii) $C_2H_6 + Br_2 \rightarrow C_2H_5Br + HBr$

(b)(iii) Substitution.

(c)
M1: (Ethane) has only single bonds / no double or triple bonds.
M2: (Therefore) it cannot add more atoms / it can only undergo substitution reactions / it contains the maximum number of hydrogen atoms per carbon.

(d)
M1 (Ethane): Bromine water stays orange/yellow / no colour change.
M2 (Ethene): Bromine water changes from orange/yellow to colourless / is decolourised.

(e)
M1: Alkenes have the general formula $C_nH_{2n}$ / there are twice as many H atoms as C atoms.
M2: Therefore, for any straight-chain alkene, the empirical formula is always $CH_2$ (when simplified by dividing by n).
M3: Alkanes have the general formula $C_nH_{2n+2}$. For $C_2H_6$, the empirical formula is $CH_3$; for $C_4H_{10}$, it is $C_2H_5$. These are different, and the “+2” means you cannot divide by n to get a constant empirical formula.

(f)
Assume 100 g of compound, so masses are: C = 19.2 g, H = 4.0 g, O = 12.8 g, Br = 64.0 g.
Divide by atomic masses (Ar):
C: $ \frac{19.2}{12} = 1.6 $
H: $ \frac{4.0}{1} = 4.0 $
O: $ \frac{12.8}{16} = 0.8 $
Br: $ \frac{64.0}{80} = 0.8 $
Divide by smallest (0.8):
C: $ \frac{1.6}{0.8} = 2 $
H: $ \frac{4.0}{0.8} = 5 $
O: $ \frac{0.8}{0.8} = 1 $
Br: $ \frac{0.8}{0.8} = 1 $
Empirical formula: $ C_2H_5OBr $
(Note: Often written as $ C_2H_5BrO $, the order does not matter for the formula).

Question 6

A student uses this apparatus to investigate the decomposition of hydrogen peroxide solution.

This is the equation for the reaction.

\[2H_2O_2 \rightarrow 2H_2O + O_2\]

(a) Give the test for oxygen.

(b) Complete the dot-and-cross diagram for a molecule of hydrogen peroxide. Show outer electrons only.

(c) The student measures the volume of oxygen collected at regular intervals until the reaction stops.

The table shows the student’s results.

Time in seconds	0	20	40	60	80	100	120	140	160
Volume of oxygen in cm$^3$	0	24	44	62	78	88	94	94	94

(i) Plot the student’s results on the grid.

(ii) Draw a curve of best fit.

(d) (i) Explain in terms of particle collision theory how decreasing the concentration affects the rate of a reaction.

(ii) The student repeats the experiment using the same volume of hydrogen peroxide solution but with half the original concentration. All other conditions are kept the same. On the grid, draw the curve you would expect the student to obtain.

(e) In this reaction, the manganese(IV) oxide acts as a catalyst. Explain how a catalyst works.

Most-appropriate topic codes (Edexcel IGCSE Chemistry 4CH1):

• 2(h): Chemical tests — part (a)
• 1(g): Covalent bonding — part (b)
• 3(b): Rates of reaction — parts (c), (d)
• 3(b): Rates of reaction (Catalysis) — part (e)

▶️ Answer/Explanation

Solution

(a)
Relights/ignites a glowing splint. (1)

(b)
M1: 3 bond pairs correct (between O-O and O-H atoms).
M2: Rest of molecule fully correct (including lone pairs on oxygen atoms).
Structure: H-O-O-H with correct electron arrangement. (2)

(c)(i) & (ii)
Graph requirements:
• All points plotted accurately to within half a small square.
• A smooth curve of best fit drawn, starting at the origin (0,0) and levelling off at 94 cm³.
(2)

(d)(i)
An explanation linking the following three points:
M1: Decreasing concentration means there are fewer reactant particles (in the same volume).
M2: This leads to fewer collisions per unit time / less frequent collisions between reactant particles.
M3: Therefore, the rate of reaction decreases. (3)
Marking note: Do not credit answers that incorrectly mention changes in kinetic energy or particle movement.

(d)(ii)
Graph requirements on the same grid:
M1: A curve starting at the origin (0,0) that is less steep (lower initial gradient) than the original curve.
M2: The curve levels off at a final volume between 46–48 cm³ (half of the original final volume, as concentration is halved). (2)

(e)
An explanation linking the following two points:
M1: A catalyst provides an alternative pathway/route for the reaction.
M2: This alternative pathway has a lower activation energy. (2)

Question 7

(a) A technician needs to identify the solution in an unlabelled bottle. The technician knows that the solution contains one of these compounds:

iron(II) bromide
iron(II) chloride
iron(III) bromide
iron(III) chloride

Describe how the technician can use chemical tests to identify the compound.

(b) When chlorine gas is passed over heated iron powder, iron(III) chloride forms. This is the equation for the reaction:

\[ 2Fe + 3Cl_2 \rightarrow 2FeCl_3 \]

0.060 mol of chlorine gas is passed over 2.8 g of iron powder. Show by calculation that the iron powder is in excess.

(c) When iron(III) chloride dissolves in water, an acidic solution forms.

(i) Give the colour of litmus in this solution.

(ii) Give the formula of the ion that causes the solution to be acidic.

Most-appropriate topic codes (Edexcel IGCSE Chemistry):

• 2(h): Chemical tests — part (a), part (c)(i)
• 2.47: Describe tests for cations including Fe$^{2+}$ and Fe$^{3+}$ — part (a)
• 2.48: Describe tests for anions including Cl$^-$ and Br$^-$ — part (a)
• 1(e): Chemical formulae, equations and calculations — part (b)
• 1.28/1.29: Calculations involving amount of substance — part (b)
• 2(f)/(g): Acids, alkalis, acids, bases and salt preparations — part (c)
• 2.31: Know that acids are a source of hydrogen ions — part (c)(ii)

▶️ Answer/Explanation

Solution

(a) A description that refers to the following six points:

M1: add sodium hydroxide (solution) or aqueous ammonia
M2: if a green precipitate forms it is an iron(II) / Fe$^{2+}$ compound
M3: if a brown precipitate forms it is an iron(III) / Fe$^{3+}$ compound
M4: add silver nitrate (solution to a fresh sample)
M5: if cream precipitate forms it is a bromide / Br$^{-}$
M6: if white precipitate forms it is a chloride / Cl$^{-}$

(b) A calculation that refers to the following points:

M1: moles of iron = $ \frac{2.8}{56} = 0.05 $ mol
M2: 2 mol iron reacts with 3 mol chlorine (from the equation)
M3: moles of Fe to react with 0.060 mol Cl$_2$ = $ 0.060 \times \frac{2}{3} = 0.04 $ mol OR mass of Fe needed = $ 0.04 \times 56 = 2.24 \, \text{g} $
Conclusion: Since available Fe (0.05 mol or 2.8 g) > required Fe (0.04 mol or 2.24 g), the iron powder is in excess.

(c)(i) red (or pink)

(c)(ii) H$^{+}$ (or H$_3$O$^{+}$)

Question 8

Diamond and graphite are made of carbon atoms, joined together by covalent bonds. The diagram shows their structures.

(a) State, in terms of electrostatic attractions, what is meant by a covalent bond.

(b) Explain why diamond has a high melting point.

(c) Explain why graphite is a good conductor of electricity.

(d) $ C_{60} $ fullerene is a molecule made of 60 carbon atoms. The diagram shows the structure of $ C_{60} $ fullerene.

One mole of atoms contains $ 6.0 \times 10^{23} $ atoms.

Determine the number of atoms in one mole of $ C_{60} $ fullerene. Give your answer in standard form.

Most-appropriate topic codes (Edexcel IGCSE Chemistry):

• 1(g): Covalent bonding — parts (a), (b), (c)
• 1(g): (Specification point 1.45) — Understand covalent bonds in terms of electrostatic attractions — part (a)
• 1(g): (Specification point 1.49) — Explain why substances with giant covalent structures are solids with high melting and boiling points — part (b)
• 1(g): (Specification point 1.50) — Explain how the structures of diamond, graphite and $C_{60}$ fullerene influence their physical properties, including electrical conductivity and hardness — parts (b), (c), (d)
• 1(e): (Specification point 1.27) — Know that the mole (mol) is the unit for the amount of a substance — part (d)

▶️ Answer/Explanation

Solution

(a)
• M1: (Electrostatic) attraction between (two) nuclei (1)
• M2: and shared/bonding pair(s) of electrons (1)
OR in reverse order: M1: attraction between shared electrons and M2: (two) nuclei.
Nuclei must be plural.

(b) An explanation that links the following three points:
• M1: Diamond is a giant covalent structure/giant lattice structure (1)
• M2: There are (many) strong covalent bonds (which need to be broken) (1)
• M3: Large amount of (heat/thermal) energy needed to break the covalent bonds (1)
Ignore “giant molecule”. No marks for reference to intermolecular forces or ions in diamond.

(c) An explanation that links the following two points:
• M1: (Graphite has) delocalised electrons (1)
• M2: (Electrons) are mobile/move/flow/can carry charge (1)
M2 is dependent on mention of electrons. Ignore “carry charge”. Zero marks if reference to ions or atoms moving.

(d) Calculation:
• M1: (Number of atoms =) $60 \times 6.0 \times 10^{23}$ (1)
• M2: $3.6 \times 10^{25}$ (1)
Correct answer without working scores 2. Answer must be in correct standard form to 1 decimal place.

Question 9

A scientist uses this apparatus in an experiment to reduce a metal oxide to a metal.

Before heating the mass of the empty tube and the mass of the tube and the metal oxide are recorded.

After heating, the tube is allowed to cool and the mass of the tube and its contents is recorded again.

(a) (i) State why the reaction of the metal oxide to form a metal is described as a reduction reaction.
(ii) State why it is important to relight the hydrogen at the end of the tube if the flame goes out.
(iii) Explain why it is important to continue passing hydrogen into the tube and burning the hydrogen at the end of the tube until the contents have cooled.
(iv) Describe what should be done next to ensure that all the metal oxide has been converted into the metal.

(b) In this experiment a mass of 4.14 g of metal is formed from 4.46 g of the metal oxide.

(i) Calculate the amount, in moles, of oxygen atoms in the sample of the metal oxide.
(ii) The formula of the metal oxide is MO, where M represents the symbol of the metal. Deduce the amount, in moles, of M in the sample of the metal oxide.
(iii) Calculate the relative atomic mass of M.
(iv) Use the Periodic Table to identify metal M.

Most-appropriate topic codes (Edexcel IGCSE Chemistry 4CH1):

• 2(d): Reactivity series — Reduction of metal oxides (Specification points 2.16, 2.20)
• 1(e): Chemical formulae, equations and calculations — Mole calculations and relative atomic mass (Specification points 1.27, 1.28, 1.33)
• 1(g): Covalent bonding — Hydrogen as a flammable gas (implied in safety)

▶️ Answer/Explanation

Solution

(a)(i) The metal oxide loses oxygen / Oxygen is removed.

(a)(ii) The gas (hydrogen) is flammable / it could cause a fire or explosion if it builds up.

(a)(iii)
• To stop oxygen/air from entering the tube. (1)
• (Because) if air entered, some of the hot metal would be re-oxidised back to the metal oxide. (1)

(a)(iv)
• Reheat the tube and its contents, allow to cool, and reweigh. (1)
• Repeat this process until a constant mass is obtained. (1)

(b)(i)
Mass of oxygen = $4.46 \, \text{g} – 4.14 \, \text{g} = 0.32 \, \text{g}$ (1)
Moles of oxygen atoms = $\frac{0.32}{16} = 0.02 \, \text{mol}$ (1)

(b)(ii)
Since the formula is MO, moles of M = moles of O = $0.02 \, \text{mol}$. (1)

(b)(iii)
Mass of M = $4.14 \, \text{g}$
Relative atomic mass, $A_r$ of M = $\frac{\text{mass}}{\text{moles}} = \frac{4.14}{0.02} = 207$ (2)

(b)(iv)
M = Lead (Pb) (1)

Question 10

This method is used in an experiment to produce hydrated zinc nitrate crystals.

pour 50 cm$^3$ of dilute nitric acid into a beaker
add a spatula of zinc powder to the acid
add more zinc until it is in excess
filter the mixture
obtain crystals of zinc nitrate from the filtrate

(a) State why the mixture is filtered.

(b) Describe how a pure, dry sample of hydrated zinc nitrate crystals could be obtained from the filtrate.

(c) This equation represents the formation of hydrated zinc nitrate in the experiment.

\[ \mathrm{Zn(s) + 2HNO_3(aq) + 6H_2O(l) \rightarrow Zn(NO_3)_2.6H_2O(aq) + H_2(g)} \]

(i) In another experiment, 9.75 g of zinc is completely reacted with nitric acid.

Show that the maximum possible mass of hydrated zinc nitrate crystals that could be formed is approximately 45 g.
\[ [\text{for } \mathrm{Zn(NO_3)_2.6H_2O}, \, M_r = 297] \] (2)

(ii) The actual yield of hydrated zinc nitrate crystals is 36.4 g.

Calculate the percentage yield of hydrated zinc nitrate crystals. (2)

Most-appropriate topic codes (Pearson Edexcel IGCSE Chemistry 4CH1):

• 2(g): Acids, bases and salt preparations — parts (a), (b) — separation and purification techniques
• 2(f): Acids, alkalis and titrations — part (a) — neutralisation reaction
• 1(e): Chemical formulae, equations and calculations — parts (c)(i), (c)(ii) — reacting masses, percentage yield
• 1(e): Chemical formulae, equations and calculations (Calculations) — part (c)(i) — use of $M_r$ and mole calculations

▶️ Answer/Explanation

Solution

(a)

To remove excess / unreacted / undissolved / insoluble zinc / solid / metal.

(b)

M1: Heat the solution to evaporate some of the water / to form a saturated solution / to crystallisation point.

M2: Leave the solution to cool / leave the solution for (more) crystals to form.

M3: Filter off the crystals.

M4: Suitable method of drying the crystals (e.g., dry between filter papers / dry in a warm oven / leave to air dry).

(c)(i)

M1: Moles of zinc = $\frac{9.75}{65} = 0.15$ mol

M2: Mass of $\mathrm{Zn(NO_3)_2.6H_2O} = 0.15 \times 297 = 44.55\ \mathrm{g} \approx 45\ \mathrm{g}$

(c)(ii)

M1: Percentage yield = $\frac{36.4}{44.55} \times 100$

M2: $= 81.7\%$

(Total = 9 marks)

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