Y13 Prep Activities For Synoptic

Question 1

This question is about acid–base titrations.
Citric acid reacts with sodium hydroxide.
C6H8O7(aq) + 3 NaOH(aq) → Na3C6H5O7(aq) + 3 H2O(l)
(a)  A student makes a solution of citric acid by dissolving some solid citric acid in water.
Describe a method to add an accurately known mass of solid to a beaker to make a solution.
(3)

Answer 1

measure the mass of the weighing boat (or similar) and solid
1
M2 Add the solid to a beaker (or other suitable container) and then reweigh the weighing boat (and subtract to find the mass of solid added.)
1
OR
M1 Place weighing boat on a balance and zero the balance
M2 Add the solid to a beaker (or other suitable container), wash out weighing boat and transfer washing to the beaker.

Question 2

The student uses this method to complete a titration.
•   Rinse a burette with distilled water.
•   Fill the burette with sodium hydroxide solution.
•   Use a measuring cylinder to transfer 25 cm3 of the citric acid solution into a conical flask.
•   Add 5 cm3 of indicator.
•   Slowly add the sodium hydroxide solution from the burette into the conical flask.
•   Add the sodium hydroxide solution dropwise near the end point until the indicator just changes colour.
•   Repeat the titration to get concordant results.
The method used by the student includes three practical steps that will lead to an inaccurate final result.
For each of these three steps
•   identify the mistake
•   explain why it is a mistake
•   suggest how the mistake can be overcome.
(6)

Answer 2

Stage 1
a. Problem – using a measuring cylinder
b. Explanation – large uncertainty / not accurate enough
c. Improvement – use a (volumetric) pipette (Not dropping pipette)
Stage 2
a. Problem – too much indicator
b. Explanation – may react and affect the endpoint reading
c. Improvement – use a smaller volume (2-6 drops)
Stage 3
a. Problem – rinsing the burette with distilled or deionised water
b. Explanation – will slightly dilute the alkali solution
c. Improvement – rinse the burette with alkali solution
Stage 4
a. Problem – adding alkali solution until the indicator “just” changes colour
b. Explanation – acid may not have fully reacted (as mixture not swirled)
c. Improvement – add alkali solution until a permanent colour change is seen.

Question 3

The total uncertainty in the use of the burette is ±0.15 cm3
Calculate the percentage uncertainty in the use of the burette in Run 1 which is 22.95

Answer 3

  (0.15 / 22.95) × 100 = 0.65%
0.15 / (Their Run 1) × 100

Question 4

The student dissolves 0.834 g of citric acid in water and makes the solution up to 500 cm3
Calculate the concentration, in mol dm–3, of citric acid in this solution.

C6H8O7(aq) + 3 NaOH(aq) → Na3C6H5O7(aq) + 3 H2O(l)

Answer 4

Mr citric acid = 192.0
1
M2 Amount of citric acid = Mass / Mr
            = 0.834 / 192
            = 0.0043438 (mol)
M2 conseq on M1
1
M3 Concentration = moles / volume
         = 0.0043438 / 0.5
         = 0.00869 (mol dm–3)

Question 5

practical 5 distillation
1.Why do we add anti bumping granules (1)
2a.Which way does water flow in a condenser (1)
b.why (1)
3a) is flask closed with a bung or not (1)
3b)And why (1)

Answer 5

1)to create smaller bubbles / to prevent large bubbles / to prevent mixture jumping into condenser

3a yes, system should be closed (above flask) to prevent gases escaping
with (bung +) thermometer
3b) to prevent distillation of other components / to stay in suitable temperature range

2a) water in from the bottom and out from the top
2b) condenser not cool enough / not full of water product may not condense / comes through as gas

Question 6

Another student completes the experiment using apparatus that is set up correctly.
(b)  The student reacts 2.0 cm3 of propan-2-ol (CH3CH(OH)CH3) with an excess of acidified potassium dichromate(VI).
The student obtains 0.954 g of propanone (CH3COCH3).
Calculate the percentage yield of propanone in this experiment.
Give your answer to the appropriate number of significant figures.
Density of propan-2-ol = 0.786 g cm–3

Answer 6

mass of propan-2-ol = 2.0 x 0.786 (= 1.572 g to at least 2sf)
M2 amount of propan-2-ol = ( = 0.0262 to at least 2 sf) mol
M3 mass of propanone expected = 0.0262 x 58.0 ( = 1.52 g to at least 2sf)
M4 % yield = = 63% (2sf only)
Alternative for M3/4
M3 amount of propanone formed = ( = 0.0164) mol
M4 % yield = = 63% (2sf only)

Question 7

This question is about gas volumes.
(a)  TNT (C7H5N3O6) is an explosive because it can decompose very quickly and exothermically to form a large volume of gas. An equation for this decomposition is
2 C7H5N3O6(s) → 3 N2(g) + 5 H2(g) + 12 CO(g) + 2 C(s)
Calculate the volume of gas, in m3, measured at 1250 °C and 101 000 Pa, produced by the decomposition of 1.00 kg of TNT (Mr = 227.0).
The gas constant, R = 8.31 J mol–1 K–1

Answer 7

amount of TNT = (= 4.41 mol)
1
M2 amount of gases formed = 10 × M1 (= 44.1 mol)
1
M3 V =
1
M4 V = converting T to 1523 (K) (or 273 + 1250)
1
M5 V = = 5.52 (m3) range 5.5(1) tp 5.53 (m3)
1
Final answer should be at least 2sf
Correct final answer scores 5 marks
Allow ECF from M1 to M2, M2 to M5, M4 to M5 and M3 to M5
0.552 (m3) for using 4.41 mol in M5 scores 4 marks (loses M2)
4.54 (m3) for using 1250 K scores 4 marks (loses M4)
3.54 (m3) for using (1250 − 273) K scores 4 marks (loses M4)
0.18 (m3) for inverted expression scores 4 marks (loses M3 or M5)
M3 can score from a substituted expression

Question 8

Molecules of propan-2-ol and propanone each contain three carbon atoms.
suggest the shape and a bond angle around the central C atom in a molecule of each compound.

(d)  Explain why propanone has a lower boiling point than propan-2-ol.

Answer 8

propan-2-ol: tetrahedral and 109.5°
M1 allow 104–110°
1
M2 propanone: trigonal planar and 120°
M2 allow 115–123°
Any two correct boxes scores one mark
1
(d) M1 propan-2-ol has stronger intermolecular forces
Penalise M1 and M2 for any reference to breaking covalent bonds, (but M3 could score)
1
M2 propan-2-ol has hydrogen bonds between molecules
For M2 ignore reference to dipole-dipole forces in propan-2-ol
1
M3 propanone has dipole-dipole forces and/or van der Waals’ forces

Question 9

)  Alkenes have the general formula CnH2n
When alkenes undergo complete combustion, 1.0 mol of CnH2n reacts with mol of oxygen.
Calculate the volume of oxygen needed for the complete combustion of 200 cm3 of but-1-ene.
The volumes of all gases are measured at the same temperature and pressure.

Answer 9

1200 (cm3)
200 ×  where n = 4 200 ×

Question 10

Alkanes have the general formula CnH2n+2
Alkanes undergo complete combustion in a plentiful supply of oxygen.
CnH2n+2 + x O2 → n CO2 + (n+1) H2O
Determine x in terms of n

Answer 10

3n+1/2
1.5n+0.5

Question 11

Ethanedioic acid (H2C2O4) is a diprotic acid. Beekeepers use a solution of this acid as a pesticide.
A student carried out a titration with sodium hydroxide solution to determine the mass of the acid in the solution. The student repeated the titration until concordant titres were obtained.
H2C2O4(aq) + 2NaOH(aq) ⟶ Na2C2O4(aq) + 2H2O(l)
(a) The student found that 25.0 cm3 of the ethanedioic acid solution reacted completely with 25.30 cm3 of 0.500 mol dm−3 sodium hydroxide solution.
Calculate the mass, in mg, of the acid in 25.0 cm3 of this solution.

Answer 11

Amount NaOH = 0.02530 × 0.500 = 0.01265 mol
567‑590 = 4 marks
0.567‑0.590 = 3 marks
1
M2 Amount acid = 0.006325 mol (i.e. M1 ÷ 2)
Allow ECF at each stage
1
M3 Mr = 90(.0)
M3 can be scored from use of value of 90(.0) within working
1
M4 mass acid = 569 (mg) (allow 567 to 576) (i.e. M2 × M3 in mg)
M4 should be to at least 2sf. Any individual marks for M1/2/3 should be to at least 2sf (or 90 for M3)
1
1134‑1180 = 3 marks (due to not dividing moles of NaOH by 2)
1.134‑1.180 = 2 marks (due to not dividing moles of NaOH by 2 and not converting to mg)

Question 12

An equation for the complete combustion of gaseous propan-1-ol is shown.
CH3CH2CH2OH+4.5O2–>3CO2+4H2O
ĀH=-1893

Table 2 shows some bond enthalpy data.

Table 2
C–H C–O O–H C=O O=O
412 360 463 805 496
Use data from Table 2 and the enthalpy change for this reaction to calculate a value for the bond enthalpy of a C–C bond in propan-1-ol.

Answer 12

∆H/-1893 = ΣB(reactants) - ΣB(products)
OR
∆H/-1893 = ΣBonds broken - ΣBonds formed
OR
∆H/-1893 = 2B(C-C) + 7B(C-H) + B(C-O) + B(O-H) + 4½ B(O=O) – 6B(C=O) – 8B(O-H)
Correct answer scores 3
1
-1893= 2B(C-C) + 7(412) + 360 + 463 + 4½(496) – 6(805) – 8(463)
OR
-1893= 2B(C-C) +5939 – 8534
OR
-1893= 2B(C-C) -2595
OR
2B(C-C) = 702

Question 13

Aqueous metal ions can be identified by test-tube reactions.
For each of the following, describe what you would observe.
Write an equation or equations for any reactions that occur.
(a) The addition of aqueous sodium carbonate to a solution containing
[Fe(H2O)6]3+(aq) ions.

Answer 13

Brown ppt/solid
1
Gas evolved/effervescence
1
2[Fe(H2O)6]3+ + 3CO32– → 2Fe(H2O)3(OH)3 + 3CO2 + 3H2O

Question 14

The addition of aqueous sodium hydroxide, dropwise until in excess, to a solution containing [Al(H2O)6]3+(aq) ions.

Answer 14

White ppt/solid
1
Colourless Solution
Only award M2 if M1 given or initial ppt mentioned
1
[Al(H2O)6]3+ + 3OH– → Al(H2O)3(OH)3 + 3H2O
Allow [Al(H2O)6]3+ + 3OH– → Al(OH)3 + 6H2O
1
Al(H2O)3(OH)3 + 3OH– → [Al(OH)6]3– + 3H2O

Question 15

The addition of concentrated hydrochloric acid, dropwise until in excess, to a solution containing [Cu(H2O)6]2+(aq) ions.

Answer 15

Green/yellow solution
1
[Cu(H2O)6]2+ + 4Cl– → [CuCl4]2– + 6H2O

Question 16

This question concerns the oxidation of a primary alcohol.
The experiment was carried out using the distillation apparatus shown in the diagram.
The oxidation product was distilled off as soon as it was formed.
Give one reason why using a water bath is better than direct heating with a Bunsen burner.

Answer 16

The alcohol is flammable

Question 17

State a simple chemical test that distinguishes the propanoic acid from the propan-1-ol.
Give one observation for the test with each substanc

Answer 17

add sodium carbonate/hydrogencarbonate
1
M2 effervescence / bubbles
not gives off (CO2) gas
1
M3 no (visible) change/reaction
not nothing / no observation
allow acidified sodium/potassium dichromate
no visible change / stays orange
orange to green

Question 18

Hydrogen peroxide decomposes to form water and oxygen.
Give an equation for this reaction.
Calculate the amount, in moles, of hydrogen peroxide that would be needed to produce 185 cm3 of oxygen gas at 100 kPa and 298 K
The gas constant, R = 8.31 J K−1 mol−1
Equation

Answer 18

2H2O2 → 2H2O + O2
allow multiples
ignore state symbols
1
M2 V = 185 × 10−6 (m3) and P = 100 000(Pa)
unit conversions
1
M3 n =
rearrangement of ideal gas equation
1
M4 n(O2) = 7.47 × 10−3 (mol)
calculation
1
M5 n(H2O2) = (7.47 × 10−3 × 2) = 0.0149 mol
allow M4 × 2 to 2 sig fig or more

Question 19

Define the term mean bond enthalpy.

Answer 19

enthalpy (change) to break 1 mol bonds (in gaseous state)
allow heat energy (change) to break 1 mol bonds
allow the enthalpy needed to break 1 mol bonds
do not accept enthalpy released
1
M2 averaged over a range of compounds / molecules

Question 20

Explain why an aqueous solution containing [Fe(H2O)6]3+ ions has a lower pH than an aqueous solution containing [Fe(H2O)6]2+ ions.

Answer 20

M1 Fe3+ is smaller (than Fe2+) OR Fe3+ has a greater charge OR Fe3+ has a greater charge density OR Fe3+ has a greater charge to size ratio
Penalise Fe(H2O)63+ ions once in M1 or M2
1
M2 Fe3+ ions are more polarising OR Fe3+ ions polarise water molecules more
1
M3 So more O-H bonds (in the water ligands) break OR more H+ ions released OR weaken O-H bonds in ligands more (in the Fe3+ solution)
Do not allow Fe3+ releases 3H+ ions

Question 21

functional group isomer of alcohol

Answer 21

ether r-0-r