dec 2017 exam paper paper 1 physical

Question 1

(d) Calculate the pH of a 0.0131 mol dm−3 solution of calcium hydroxide at 10 °C

Give your answer to two decimal places.

Answer 1

(d) [OH−] = 0.0131 × 2 = 0.0262
[H+] = (Kw / [OH−] ) = 2.93 × 10−15 / 0.0262 (= 1.118 × 10−13)

1

pH = (− log (1.118 × 10−13) = 12.9514 = 12.95

Or

[OH−] = 0.0131 × 2 = 0.0262

pOH = (−log 0.0262) = 1.5817

pH =(−log Kw− pOH = −log 2.93 × 10−15−1.58 = 14.53 − 1.58) = 12.95

allow to 2dp or more

Question 2

(c) Suggest why this pure water at 10 °C is not alkaline.

Answer 2

(c) [H+] = [OH-]

Question 3

(e) The 0.0131 mol dm−3 calcium hydroxide solution at 10 °C was a saturated solution.

A student added 0.0131 mol of magnesium hydroxide to 1.00 dm3 of water at 10 °C and stirred the mixture until no more solid dissolved.

Predict whether the pH of the magnesium hydroxide solution formed at 10 °C is larger than, smaller than or the same as the pH of the calcium hydroxide solution at 10 °C

Explain your answer.

Answer 3

(e) smaller / lower pH / less alkaline / more acidic

If not smaller CE = 0/2

Allow pH number between 8 and 12

1

(magnesium hydroxide) is less soluble / sparingly soluble/ solubility of hydroxide increases down group II

M2 dependent on M1 but if blank mark on

Ignore concentration and dissociation

Ignore incorrect formula

Do not allow Mg(OH)2 is insoluble

Question 4

(b) Write an equation, including state symbols, to show how an atom of titanium is ionised by electron impact and give the m/z value of the ion that would reach the detector first.
(c) Calculate the mass, in kg, of one atom of 49Ti

The Avogadro constant L = 6.022 × 1023 mol−1

Answer 4

(b) Ti(g) → Ti+(g) +e−

or Ti(g) + e−→ Ti+(g) +2e−

or Ti(g) − e−→ Ti+(g)

State symbols essential

Allow electrons without − charge shown.
c)8.1(37) × 10−26

Question 5

(d) In a TOF mass spectrometer the time of flight, t, of an ion is shown by the equation

In this equation d is the length of the flight tube, m is the mass, in kg, of an ion and E is the kinetic energy of the ions.

In this spectrometer, the kinetic energy of an ion in the flight tube is 1.013 × 10−13 J

The time of flight of a 49Ti+ ion is 9.816 × 10−7 s

Calculate the time of flight of the 47Ti+ ion.

Give your answer to the appropriate number of significant figures.

Answer 5

(d) M1 is for re-arranging the equation

Allow t α square root of m

1

Or

d = 1.5(47)

This scores 2 marks

Allow this expression for M2

1

= 9.6(14) × 10−7

Question 6

This question is about some Period 3 elements and their oxides.

(a) Write an equation for the reaction of phosphorus with an excess of oxygen.
b) Describe a test you could carry out in a test tube to distinguish between sodium oxide and the product of the reaction in part (a)

Answer 6

(a) P4 + 5 O2 → P4O10
b) React with water / add water / solution (of substances in question)

Add litmus paper / universal indicator / measure pH (with pH meter)
Allow other reagents in solution, e.g. sodium carbonate solution, that give a positive result
M3:Litmus: blue with sodium oxide (solution) and red with phosphorus oxide (solution)OR

If blue litmus added phosphorus oxide solution goes red OR

If red litmus added sodium (hydr)oxide goes blue

Universal Indicator: blue/ purple with sodium oxide (solution) and red with phosphorus oxide (solution)

pH meter or Universal Indicator: sodium (hydr)oxide (solution) has a higher pH (than phosphorus oxide (solution)) or vv

sodium (hydr)oxide pH (12 to 14) and phosphorus oxide (solution) pH (-1 to 2)

For pH meter or Universal Indicator: allow sodium (hydr)oxide (solution) has a higher pH and phosphorus oxide

Question 7

(c) State the type of crystal structure shown in silicon dioxide and in sulfur trioxide.

Answer 7

(c) For silicon dioxide - giant covalent (molecule)/ macromolecular

1

For sulfur trioxide - molecular / (simple) molecule

1

Do not allow simple covalent

Question 8

(d) Explain why silicon dioxide has a higher melting point than sulfur trioxide.

Answer 8

(d) Covalent bonds (between atoms) in SiO2

1

Van der Waals between molecules / intermolecular forces in SO3

1

Covalent bonds are stronger than van der Waals forces

1

(Covalent bonds) take more energy to be overcome/broken or (Van der Waals) take less energy to be overcome/broken

1

If covalent bonds between molecules of SiO2 lose M1 only

If hydrogen bonds in SO3 lose M2 only

If metallic or ionic max score = 1 (either M1 or M2)

Question 9

(a) Describe how a student could distinguish between aqueous solutions of potassium nitrate, KNO3, and potassium sulfate, K2SO4, using one simple test-tube reaction.

Answer 9

(a) BaCl2 / Ba(OH)2 / Ba(NO3)2 / BaX2 or names

Question 10

(f) Write an equation for the reaction of an excess of magnesium oxide with phosphoric acid.

Answer 10

(f) 3 MgO + 2 H3PO4 → Mg3(PO4)2 + 3 H2O

Question 11

(e) Write an equation for the reaction of sulfur trioxide with potassium hydroxide solution.

Answer 11

(e) SO3 + 2KOH → K2SO4 + H2O

SO3 + KOH → KHSO4

SO3 + 2OH− → SO42−+ H2O

SO3 + OH− → HSO4−

Allow multiples

Question 12

(b) Describe how a student could distinguish between aqueous solutions of magnesium chloride, MgCl2, and aluminium chloride, AlCl3, using one simple test-tube reaction.

Answer 12

(b) NaOH / sodium hydroxide / other Group 1 hydroxides

If reagent incorrect or blank then CE =0/3

If reagent incomplete, lose M1 and mark on

1

white precipitate / white solid

1

(white) ppt which dissolves in excess (NaOH)

If reagent is excess NaOH, allow colourless solution for M3
Alternative Method

Name or formula of Group 1 carbonate

1

white precipitate / white solid

1

(white) precipitate and effervescence

Question 13

(b) The equation shows the final stage in the production of methanol.

CO(g) + 2H2(g) ⇌ CH3OH(g)

20.1  mol of carbon monoxide and 24.2  mol of hydrogen were placed in a sealed container. An equilibrium was established at 600  K. The equilibrium mixture contained 2.16  mol of methanol.

Calculate the amount, in moles, of carbon monoxide and of hydrogen in the equilibrium mixture. 2

Answer 13

(b) Moles of carbon monoxide 17.9

Allow 17.94

1

Moles of hydrogen 19.9

Allow 19.88

Question 14

(c)
CO(g) + 2H2(g) ⇌ CH3OH(g) A different mixture of carbon monoxide and hydrogen was allowed to reach equilibrium at 600 K

At equilibrium, the mixture contained 2.76  mol of carbon monoxide, 4.51  mol of hydrogen and 0.360  mol of methanol. The total pressure was 630  kPa

Calculate a value for the equilibrium constant, Kp, for this reaction at 600  K and state its units. 6

Answer 14

If Kp expression incorrect can only score M2 & M3 & M4
pp(CH3OH)/pp(CO) pp(H2)2
pp(CO)=228
pp(H2)2=372
pp(CH3OH)=29.7
29.7 / 228x(372x2)=9.4x10^-7
kPa-2

Question 15

This question is about compounds containing ethanedioate ions.

(a) A white solid is a mixture of sodium ethanedioate (Na2C2O4), ethanedioic acid dihydrate (H2C2O4.2H2O) and an inert solid. A volumetric flask contained 1.90  g of this solid mixture in 250  cm3 of aqueous solution.

Two different titrations were carried out using this solution.

In the first titration 25.0 cm3 of the solution were added to an excess of sulfuric acid in a conical flask. The flask and contents were heated to 60 °C and then titrated with a 0.0200 mol dm−3 solution of potassium manganate(VII). When 26.50 cm3 of potassium manganate(VII) had been added the solution changed colour.

The equation for this reaction is

2MnO4− + 5C2O42− + 16H+ → 2Mn2+ + 8H2O + 10CO2

In the second titration 25.0 cm3 of the solution were titrated with a 0.100 mol dm−3 solution of sodium hydroxide using phenolphthalein as an indicator. The indicator changed colour after the addition of 10.45 cm3 of sodium hydroxide solution.

The equation for this reaction is

H2C2O4 + 2OH− → C2O42− + 2H2O

Calculate the percentage by mass of sodium ethanedioate in the white solid.

Give your answer to the appropriate number of significant figures.

Show your working.(8)

Answer 15

(a) Moles MnO4−= 26.5x0.02 / 1000= 5.30 × 10−4
Moles in 25cm3 sample / pipette C2O42− ( from acid and salt)

Moles NaOH = 10.45x0.1 / 1000( = 1.045 × 10−3)
So moles C2O42− from acid in 25cm3 sample / pipette

= 1.045 × 10−3 ÷ 2 = 5.225 × 10−4
Hence moles C2O42− in sodium ethanedioate in 25 cm3

= 1.325 × 10−3 – 5.225 × 10−4 (= 8.025 × 10−4)
So moles C2O42− in sodium ethanedioate in original sample

= 8.025 × 10–4 × 10 (= 8.025 × 10−3)
Mass Na2C2O4 = 8.025 × 10−3 × 134(.0) = 1.075(35) g
1.075(35) / 1.9 x100 =56.6%
M8 = (M7/1.90) × 100 Allow 56.5 – 56.8%

So % sodium ethanedioate in original sample

Question 16

(b) Ethanedioate ions react with aqueous iron(III) ions in a ligand substitution reaction.

Write an equation for this reaction.

Suggest why the value of the enthalpy change for this reaction is close to zero.
d) Ethanedioate ions are poisonous because they react with iron ions in the body. Ethanedioate ions are present in foods such as broccoli and spinach.

Suggest one reason why people who eat these foods do not suffer from poisoning.

Answer 16

(b) [Fe(H2O)6]3+ + 3C2O42− → [Fe(C2O4)3]3− + 6H2O

1

There are 6 Fe–O bonds broken and then made / same number and type of bond being broken and made.
d) The ethanedioic acid is only present in small quantities/low

concentration in these foods.