This question is about sodium fluoride (NaF).

Some toothpastes contain sodium fluoride.The concentration of sodium fluoride can be expressed in parts per million (ppm).1 ppm represents a concentration of 1 mg in every 1 kg of toothpaste.

(a)     A 1.00 g sample of toothpaste was found to contain 2.88 × 10–5 mol of sodium fluoride.

Calculate the concentration of sodium fluoride, in ppm, for the sample of toothpaste.Give your answer to 3 significant figures.

Concentration of sodium fluoride _______________________________ ppm

(4)

1

(b)     Sodium fluoride is toxic in high concentrations.Major health problems can occur if concentrations of sodium fluoride are greater than 3.19× 10–2 g per kilogram of body mass.

Deduce the maximum mass of sodium fluoride, in mg, that a 75.0 kg person could swallowwithout reaching the toxic concentration.

Mass of sodium fluoride _______________________________ mg

(1)

(c)     The concentration of sodium fluoride in a prescription toothpaste is 2800 ppm.

Use your answer to Question (b) to deduce the mass of toothpaste, in kg, that a 75.0 kgperson could swallow without reaching the toxic concentration.

Mass of toothpaste _______________________________ kg

(1)

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(d)     Identify the diagram in the figure below that shows the correct relative sizes of the ions insodium fluoride.Justify your answer.

Diagram __________

Justification _________________________________________________________

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(3)

(Total 9 marks)

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A student added 627 mg of hydrated sodium carbonate (Na2CO3.xH2O) to 200 cm3 of 0.250 mol

dm–3 hydrochloric acid in a beaker and stirred the mixture.After the reaction was complete, the resulting solution was transferred to a volumetric flask,made up to 250 cm3 with deionised water and mixed thoroughly.Several 25.0 cm3 portions of the resulting solution were titrated with 0.150 mol dm–3 aqueoussodium hydroxide. The mean titre was 26.60 cm3 of aqueous sodium hydroxide.

Calculate the value of x in Na2CO3.xH2OShow your working.Give your answer as an integer.

Value of x ____________________

(Total 7 marks)

2

Some 1.0 mol dm–3 solutions were mixed using equal volumes of each solution.

Which pair of solutions would give the greatest mass of solid? 

A Ba(OH)2 and MgCl2

B Ba(OH)2 and MgSO4

C Ba(OH)2 and NaCl

D Ba(OH)2 and Na2SO4

(Total 1 mark)

3

Copper can be produced from rock that contains CuFeS2

(a)     Balance the equations for the two stages in this process.

.....CuFeS2 + .....O2 + .....SiO2  ⟶  .....Cu2S + .....Cu2O + .....SO2 + .....FeSiO3

.....Cu2S + .....Cu2O  ⟶  .....Cu + .....SO2

(2)

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(b)     Suggest two reasons why the sulfur dioxide by-product of this process is removed from theexhaust gases.

Reason 1 ___________________________________________________________

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Reason 2 ___________________________________________________________

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(2)

(c)     A passenger jet contains 4050 kg of copper wiring.

A rock sample contains 1.25% CuFeS2 by mass.

Calculate the mass, in tonnes, of rock needed to produce enough copper wire for apassenger jet.           (1 tonne = 1000 kg)

Mass of rock ____________________ tonnes

(4)

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(d)     Copper can also be produced by the reaction of carbon with copper(II) oxide according tothe equation

2CuO + C ⟶ 2Cu + CO2

Calculate the percentage atom economy for the production of copper by this process.

Give your answer to the appropriate number of significant figures.

Percentage atom economy ____________________

(2)

(Total 10 marks)

A 4.85 g sample of anhydrous sodium sulfate is dissolved in water and the solution made up to250 cm3 in a volumetric flask.

What is the concentration in mol dm−3 of sodium sulfate in the solution? 

A 0.0341

B 0.137

C 0.163

D 0.273

(Total 1 mark)

5

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A 20.0 cm3 sample of a 0.400 mol dm−3 aqueous solution of a metal bromide (MBrn) reacts

exactly with 160 cm3 of 0.100 mol dm−3 aqueous silver nitrate.

What is the formula of the metal bromide? 

A MBr

B MBr2

C MBr3

D MBr4

(Total 1 mark)

6

A sample of strontium ore is known to contain strontium oxide, strontium carbonate and someinert impurities. To determine the mass of strontium carbonate present, a student weighed asample of the solid ore and then heated it in a crucible for 5 minutes. The sample was allowed tocool and then reweighed. This heating, cooling and reweighing was carried out three times.

The results are set out in the table. 

Mass of crucible / g 9.85

Mass of crucible and ore sample / g 16.11

Mass of crucible and sample after first heating / g 14.66

Mass of crucible and sample after second heating / g 14.58

Mass of crucible and sample after third heating / g 14.58

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(a)    When strontium carbonate is heated it decomposes according to the following equation.

SrCO3 ⟶ SrO + CO2

Give a reason why the mass of the solid sample changed during the experiment.

___________________________________________________________________

___________________________________________________________________

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(1)

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(b)    Use the data in the table to calculate the mass of strontium carbonate in the original oresample. Give your answer to an appropriate precision.

Mass of strontium carbonate = ___ g

(5)

(c)    Each balance reading has an uncertainty of ±5.00 mg.

Calculate the percentage error in the initial mass of ore used.

Percentage error = ____________ %

(1)

(d)    The mass of inert impurities in the sample was 347 mg.

Deduce the mass of SrO in the sample and justify any assumption made in calculating youranswer.(If you have been unable to answer part (b), assume the mass of strontium carbonate was4.85 g. This is not the correct answer.)

Mass of SrO = _______________

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(2)

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(e)    Strontium metal can be extracted by heating strontium oxide with aluminium metal.

In this reaction, strontium vapour and solid aluminium oxide are formed.

Write an equation for the reaction and state the role of the aluminium in the process.Explain why strontium forms a vapour but aluminium oxide is formed as a solid.

Equation

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Role of aluminium ___________________________________________________

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Explanation ________________________________________________________

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(5)

(Total 14 marks)

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Mark schemes

(a)     Mr NaF = 42(.0)

Incorrect Mr loses M1 & M41

Mass NaF in 1 g = 2.88 × 10–5 × 42.0 (= 1.210 (1.2096) × 10–3 g)1

Mass NaF in 1 kg = 1.210 (1.2096) g

M3 = M2 × 1000 (g)

Units, if given, must match answer1

(Mass in mg = 1210 (1209.6) mg)

Concentration of NaF = 1210 (ppm)

Allow 1.21 × 103 ppm1

1

(b)     Toxic mass = 3.19 × 10–2 × 75 × 1000= 2390 mg

Allow 23931

(c)     Mass of toothpaste needed =                  = 0.854 kg

Mark consequential to (b)

(b) ÷ 2800 (to at least 2 sig fig)

Allow 0.85 - 0.86 kg1

(d)     B

If not B, allow M2 only

If blank, read on.1

Both Na+ and F– same electron arrangement (1s22s22p6) or isoelectronic

Electronegativity, molecules or IMF = CE, M1 only1

Sodium (ion) has more protons so attracts (outer) electrons closer / Sodium (ion) has moreprotons so stronger attractions for (outer) electrons

Ignore shielding, higher charge density, atomic radius

If reference to fluorine rather than fluoride, then penalise 1 markonly

1

[9]

M1    HCl added = 0.050 mol andNaOH used in titration = 3.99 × 10–3 mol

1

2

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M2    So moles that would be needed to neutralise total excessHCl = 3.99 × 10–3 × 10 = 3.99 × 10–2 mol

Alternative: divide moles HCl by 10 = 0.005 and 0.005 - 3.99 × 10–3

= 0.001011

M3    Therefore the moles of HCl reacted with the Na2CO3.xH2O =

0.050 - 3.99 × 10–2 = 0.0101 mol

Alternative: 0.00101 × 10 to produce 0.01011

M4    So moles Na2CO3.xH2O reacted with the HCl = 0.0101 / 2 = 5.05 x

10–3 mol1

M5    Conversion of mg to g = 0.627 (g) or 627 × 10–3 (g)1

M6    xH2O = 0.627/5.05 × 10–3 -106.0 = 18 (.16)

Alternative: mass Na2CO3 that reacted with the HCl 5.05 × 10–3

x106.0 = 0.5353 g and mass H2O = 0.627- 0.5353 = 0.0917 g1

M7    so x = 1

Alternative: 0.0917 /18.0 = 5.094 × 10–3 so ratio

Na2CO3 to H2O = 1:1.009 ie 1:1 so × = 11

[7]

B

[1]3

(a)     4CuFeS2   +   9 O2   +   4SiO2   ⟶   Cu2S   +   Cu2O   +   7SO2   +   4FeSiO3

Allow multiples1

Cu2S   +   2Cu2O   ⟶   6Cu + SO21

4

(b)     ANY TWO•   Prevents acid rain (which damages buidlings / ecology)•   Toxic OR causes breathing problems•   Reduces waste product OR makes use of the waste OR improves atom economy OR

Reduces need for sulfur mining OR used to produce sulfuric acid OR any namedproducts

2

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(c)     M1, M2, M3 are process marks

M1     Mol Cu = (= 63780)

1

M2     Mass CuFeS2 = (63780) × 183.5 (= 1.17 × 107g)1

M3     Mass ore = (1.17 × 107) × 1

M4     Mass ore = 936 tonnes (Allow 936 –937)1

Alternative methodM1 % of Cu in CuFeS2 = (63.5/183.5) × 100 = 34.6%M2 % of Cu in the rock = (34.6/100) × 1.25 = 0.4325%M3 mass of rock = 4050 × 100/0.4325 = 936416kgM4 mass of rock in tonnes = 936 tonnes

NotesM1 Ar Cu must be usedM2 Mr CuFeS2 to have been usedM3 Grossing up for the mass of rockM4 Final answer correct in tonnes

(d)     % atom economy = × 1001

= 74.3% must be 3sf1

[10]

B

[1]5

B

[1]6

(a)     CO2 gas escapes or is lost17

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(b)     Mass CO2 = 16.11 − 14.58 = 1.53 g1

Mr CO2 = 44.0

Mol CO2 = 1.53 / 44.0 = 3.48 × 10−2

1

Mol SrCO3 = 3.48 × 10−2

1

Mass SrCO3 = mol × Mr = 3.48 × 10−2 × 147.6

Mass SrCO3 = 5.13 (g)

1 mark for the answer and 1 for 3 sf precision

Allow 5.14 g (as a result of rounding)2

(c)     Percentage error =  

= 0.160 (%)1

(d) Original Mass SrO = 6.26 − 0.347 − 5.13

       = 0.783 g (or 783 mg)

OR 6.26 − 0.347 − 4.85 = 1.063 gAllow 0.773 g or 773 mg (from rounding error in part (b)

1

Justification: All SrCO3 reacted because heated to constant mass.1

(e)     3SrO + 2Al ⟶ Al2O3 + 3Sr1

Al acts as a reducing agent1

Sr is collected as a vapour because1

Al2O3 is an ionic lattice and so has strong ionic attractions1

Than Sr which is a metallic structure with (relatively) weaker bonding1

[14]

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