Year 11 Chemistry Prac Test 2015 Answers

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YEAR 11 PRACTICAL
ASSESSMENT TEST TASK 3
2015
Chemistry
Name: _________________________
Teacher: _______________________
Time allowed: 51 minutes
Total Marks: 30
Instructions
 Attempt all questions at each of three stations.
 Answer all questions in the spaces provided.
 Please leave all equipment as you found it before moving onto the next station.
 Supply your own calculator (they will not be provided).
 A periodic table will be provided for this test.
 There are 3 stations. You are allowed 17 minutes for each station.
______________________________________________
(Working area - any work in this area will not be marked)
Answers
STATION 1 Data Analysis (10 marks)
 Refer to the following graph and answer the questions below.
To convert temperature in Kelvin to
Celsius, subtract 273.15 degrees.
(a) Describe the general trend shown by this graph.
(H2O does not fit the general trend so don’t consider it when describing the general trend.)
(1 mark)
● The boiling point / temperature increases as the molecular mass increases. (1 mark)
(b) Explain the general trend shown by this graph.
(2 marks)
(H2O does not fit the general trend so don’t consider it when explaining the general trend.)
● Boiling point increases with molecular mass because dispersion forces increase with
molecular mass. (+1 mark)
● Dispersion forces increase with molecular mass because the number of electrons
increases with molecular mass. (+1 mark)
● Increasing intermolecular forces. (1 mark)
(c) Draw a Lewis Dot Diagram of the H2O molecule.
(2 marks)
● Correct including bent shape (2 marks) Accept dots and/or crosses
● Correct but not showing bent shape (1 mark)
● Correct with bent shape but not showing two lone electron pairs (1 mark)
(d) Explain why H2O does not fit the general trend shown by this graph.
(2 marks)
● States that hydrogen bonding produces very strong intermolecular forces and links this
to higher boiling point / difficulty of molecules escaping the liquid etc. (2 marks)
● States that hydrogen bonding produces very strong intermolecular forces. (1 mark)
● Mentions hydrogen bonding. (12 mark)
(e) Calculate the mass of water produced when 74.37L of methane (CH4), measured at 25°C and 100kPa,
is completely combusted.
(3 marks)
𝐶𝐻4 + 2𝑂2 → 𝐶𝑂2 + 2𝐻2 𝑂
● Writes correctly balanced equation (+1 mark)
𝑣𝐶𝐻4
74.37𝐿
𝑛𝐶𝐻4 =
=
= 3 𝑚𝑜𝑙
● Calculates 𝑛𝐶𝐻4 (+1 mark)
𝑉𝑀
24.79𝐿 ⋅ 𝑚𝑜𝑙 −1
𝑛𝐻2 𝑂 = 2𝑛𝐶𝐻4 = 6 𝑚𝑜𝑙
● Calculates 𝑛𝐻2𝑂 (+12 mark) Allow for follow-on errors.
● Calculates 𝑚𝐻2𝑂 (+12 mark) Allow for follow-on errors.
𝑚𝐻2 𝑂 = 6 𝑚𝑜𝑙 × 18.016𝑔 ⋅ 𝑚𝑜𝑙 −1 = 108.1𝑔
● No or incorrect units for final answer (- 12 mark)
OR 𝑚𝐻2 𝑂 = 6 𝑚𝑜𝑙 × 18𝑔 ⋅ 𝑚𝑜𝑙 −1 = 108𝑔
***Ignore significant figures***
STATION 2 Empirical Formula (10 marks)
 A group of students performed an experiment to determine the empirical formula of magnesium oxide
by combusting magnesium metal in a crucible. They recorded the following results.
Mass of magnesium metal (g)
Mass of empty crucible with lid (g)
Mass of crucible with lid and magnesium oxide (g)
9.72
59.91
75.74
(a) Write a balanced chemical equation for the reaction that occurred. Include all states.
● 𝑀𝑔(𝑠) + 12𝑂2(𝑔) → 𝑀𝑔𝑂(𝑠) OR 2𝑀𝑔(𝑠) + 𝑂2(𝑔) → 2𝑀𝑔𝑂(𝑠) (2 marks)
● Correctly balanced equation but without all states (1 mark)
● Correctly balanced equation based on incorrect formula for MgO (+12 mark)
(2 marks)
(b) Use the group’s results to determine the empirical formula of magnesium oxide.
(3 marks)
𝑚𝑀𝑔
𝑛𝑀𝑔 = 𝑀
𝑀𝑔
9.72𝑔
= 24.31𝑔⋅𝑚𝑜𝑙 −1 = 0.3998 𝑚𝑜𝑙
𝑚𝑂 = (75.74 − 59.91)𝑔 − 9.72𝑔 = 6.11𝑔
𝑛𝑂 =
𝑚𝑂
𝑀𝑂
6.11𝑔
= 16.00𝑔⋅𝑚𝑜𝑙−1 = 0.3819 𝑚𝑜𝑙
0.3998 0.3819
● Calculates 𝑛𝑀𝑔 (+12 mark)
● Calculates 𝑚𝑂 (+1 mark)
● Calculates 𝑛𝑂 (+12 mark) Allow for follow-on errors.
𝑛𝑀𝑔 ∶ 𝑛𝑂 = 0.3998 ∶ 0.3819 = 0.3819 : 0.3819 = 1.047: 1
● Finds the ratio (+12 mark) Allow for follow-on errors.
The empirical formula of magnesium oxide is MgO.
● States the correct empirical formula (+12 mark)
(c) Explain why the empirical formula is the same as the chemical formula for Magnesium Oxide but
generally not the same for hydrocarbons.
(2 marks)
● Salt formulas are the simplest ratio, just like empirical formulas.
(+1 mark)
● The chemical formula for molecules counts the atoms in the molecule / is not a ratio.
(+1 mark)
(d) Determine the chemical formula of Sodium Oxide using the Periodic Table.
(1 mark)
● 𝑁𝑎2 𝑂 (1 mark)
(e) One of the students observed that the empirical formula they calculated was not quite a whole number
ratio. Identify possible sources (not types) of error in their experiment to account for this. (2 marks)
● Incomplete combustion of magnesium (+1 mark)
● Loss of magnesium oxide in fumes (+1 mark)
● Inaccurate weighing (+12 mark)
Also accept valid proposals for minimising incomplete combustion and loss of magnesium
oxide in fumes.
STATION 3 Solubility (10 marks)

Perform an experiment to determine the solubility of sodium chloride (NaCl), sucrose (C12H22O11),
sand (SiO2), polyethylene (C2H4)n and decane (C10H22) in separate trials. Place 14 of a teaspoon of
sodium chloride, sucrose, sand and one piece of polyethylene into the beaker provided then add
approximately 100ml of water and stir for 10 seconds to try to dissolve. Add approximately 100ml of
water to the pre-prepared beaker containing 10ml of decane and stir for 10 seconds.
*DO NOT pour the used sand, polyethylene or decane down the sink. DISPOSE OF THESE
SUBSTANCES IN THE LARGE LABELLED CONTAINERS.
(a) Record your results in a table.
Substance
sodium chloride / NaCl
sucrose / C12H22O11
sand / SiO2
polyethylene / (C2H4)n
decane / C10H22
Soluble / Dissolves
yes / soluble
yes / soluble
no / insoluble
no / insoluble
no / insoluble
(2 marks)
● Complete and correct data (+1 mark)
● Correct but incomplete data (+12 mark)
●Well-constructed table (+1 mark)
***Ignore order of substances in table***
(b) Choose TWO of the five substances and explain why the substance was soluble or insoluble in water
in terms of bonding and/or intermolecular forces.
(2 marks)
● NaCl: Strong ion-dipole interactions break up lattice / H2O molecules stabilise free ions (+1 mark)
● Sucrose: Polar molecule in polar solvent / strong hydrogen bonding with water molecules (+1 mark)
● Sand: Covalent network bonds too strong to be broken by vibrations of water molecules (+1 mark)
● Polyethylene: Too large / non-polar molecule in polar solvent / hydrophobic (+1 mark)
● Decane: Too large / non-polar molecule in polar solvent / hydrophobic (+1 mark)
(c) 2.4L of 3.0M sodium hydroxide solution was diluted by adding pure water to produce a new solution
with a concentration of 0.90 mol.L-1. Calculate the volume of the diluted solution.
(2 marks)
−1
𝑐1 𝑣1 3.0 𝑚𝑜𝑙 ⋅ 𝐿 × 2.4𝐿
𝑣2 =
=
= 8.0𝐿
𝑐2
0.90𝑚𝑜𝑙 ⋅ 𝐿−1
● Calculates 𝑣2 with correct units and to 2 sig. fig. (2 marks)
● No or incorrect units for final answer (- 21 mark)
(d) Draw a diagram showing 1 unit of sodium chloride dissolved in water.
(2 marks)
● Both ions drawn and labelled (+1 mark)
● Both ions drawn but one or both ions not labelled (+21 mark)
● At least one water molecule per ion facing the right way (+1 mark)
(e) The maximum solubility of sodium bromide in water at 30oC is 90.8g of salt per 100ml of water. How
many moles of sodium bromide are able to dissolve in 1L of water at a 30oC?
(2 marks)
90.8g dissolves in 100ml so 908g dissolves in 1L
𝑚𝑁𝑎𝐵𝑟
908𝑔
908𝑔
𝑛𝑁𝑎𝐵𝑟 =
=
= 8.82𝑚𝑜𝑙 𝑂𝑅 =
= 8.82𝑚𝑜𝑙
−1
(23 + 80)𝑔 ⋅ 𝑚𝑜𝑙 −1
𝑀𝑁𝑎𝐵𝑟 (22.99 + 79.90)𝑔 ⋅ 𝑚𝑜𝑙
● Correctly calculates moles / solubility and includes correct units (2 marks)
(Accept 𝑚𝑜𝑙, moles, 𝑚𝑜𝑙 ⋅ 𝐿−1 and 𝑚𝑜𝑙
for units)
𝐿
● No or incorrect units for final answer. (- 21 mark)
● One error in calculation. (-1 mark)
***Ignore significant figures***
End of Test
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