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Required Practical Questions Flashcards

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1
Q

This question is about a white solid, MHCO3, that dissolves in water and reacts with hydrochloric acid to give a salt.

MHCO3 + HCl -> MCl + H2O + CO2

A student was asked to design an experiment to determine a value for the Mr of MHCO3.
The student dissolved 1464 mg of MHCO3 in water and made the solution up to 250 cm^3.
25.0 cm^3 samples of the solution were titrated with 0.102 mol dm^-3 hydrochloric acid.
The results are shown in the table.

Inital Burette ( 1 ) reading / cm^3

Final burette ( 2 ) reading / cm^3

Titre / cm^3 ( 3 )

Rough:

1 ) 0.00

2 ) 10.00

3 ) 10.00

1:

1 ) 10.00

2 ) 19.50

3 ) 9.50

2:

1 ) 19.50

2 ) 29.25

3 ) 9.75

3:

1 ) 29.25

2 ) 38.90

3 ) 9.65

Calculate the mean titre and use this to determine the amount, in moles, of HCl that reacted with 25.0 cm^3 of the MHCO3 solution.

A
  • Mean titre = ( 9.75 + 9.65 ) / 2 = 9.70 cm^3
  • Mol of HCl = 0.102 x 9.7 x 10^-3 = 9.894 x 10^-4 ( dm^-3 )
  • Mol of HCl = 9.894 x 10^-4
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2
Q

Calculate the amount, in moles, of MHCO3 in 250 cm3 of the solution.
Then calculate the experimental value for the Mr of MHCO3.
Give your answer to the appropriate number of significant figures.

( Mol of HCl = 9.894 x 10^-4 in 25 cm^3 ) ( 1 : 1 )
( Mass of MHCO3 = 1464 mg )

A
  • Mol of MHCO3 = 9.894 x 10^-4 x 10 = 9.89 x 10^-3 ( 25.0 cm^-3 -> 250 cm^-3 )
  • Mr = ( 1464 x 10^-3 ) / 9.89 x 10^-3 = 148.0283114
  • Mr = 148
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3
Q

The student identified use of the burette as the largest source of uncertainty in the experiment.
Using the same apparatus, suggest how the procedure could be improved to reduce the percentage uncertainty in using the burette.
Justify your suggested improvement.

A
  • Suggestion: Use a larger mass of solid

- Justification: therefore you get a larger titre

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4
Q

Another student is required to make up 250 cm3 of an aqueous solution that contains a known mass of MHCO3.
The student is provided with a sample bottle containing the MHCO3.
Describe the method, including apparatus and practical details, that the student should use to prepare the solution.

A
  • Weigh the sample bottle containing the MHCO3
  • Transfer the solid into a beaker and reweigh the bottle
  • Calculate the difference in mass
  • Dissolve the solid MHCO3 in water and completely dissolve it
  • Put the solution into a volumetric flask and fill it up to the 250 cm^3 miniscus, ensure that washings from the initial beaker and stirring rod is used to ensure complete transfer
  • Invert the flask to mix the solution
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5
Q

There is an experimental method for determining the number of water molecules in the formula of hydrated sodium carbonate.
This method involves heating a sample to a temperature higher than 300 °C and recording the change in mass of the sample.
The equation for the reaction taking place is:

Na2CO3.10H2O( s ) -> Na2CO3( s ) + 10H2O( g )

A group of six students carried out this experiment.
They each weighed out a sample of hydrated sodium carbonate.
They then heated their sample to a temperature higher than 300°C in a crucible for ten minutes and recorded the final mass after the crucible had cooled.
Their results are summarised in the table.

Inital mass / g ( 1 )

Final mass / g ( 2 )

( Students )

1:

1 ) 2.43

2 ) 0.90

2:

1 ) 1.65

2 ) 0.61

3:

1 ) 3.58

2 ) 1.53

4:

1 ) 1.09

2 ) 0.40

5:

1 ) 2.82

2 ) 1.15

6:

1 ) 1.95

2 ) 0.72

Plot the values of Initial mass ( y-axis ) against Final mass.
A graph of these results should include an additional point.
Draw a circle on the grid around the additional point that you should include.

A
  • Go up to 4 for initial mass
  • Go up to 2 for final mass
  • The additional point that should be included is ( 0, 0 )
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6
Q

Draw a best-fit straight line for these results that includes your additional point.

( Graph of the six students of initial and final mass )

A
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7
Q

Identify each student whose experiment gave an anomalous result.

A
  • Student 3 and 5 ( The ones with a final mass over 1g )
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8
Q

All the students carried out the experiment exactly according to this method.
Explain why a student that you identified in part ( c ) obtained an anomalous result.

( Six students question of initial mass and final mass of product )

( Students 3 and 5 were identified to be the anomalous ones )

A
  • Sample 5 hasn’t lost all their water

- as it wasn’t heated for enough time

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9
Q

A green solution, X, is thought to contain [ Fe( H2O )6 ]^2+ ions.

The presence of these ions can be confirmed by reacting separate samples of solution X with aqueous ammonia and with aqueous sodium carbonate.
Write equations for each of these reactions and describe what you would observe.

A
  • [ Fe( H2O )6 ]^2+ + 2NH3 -> Fe( H2O )4 ( OH )2 + 2NH4^+
  • Green precipitate
  • [ Fe( H2O )6 ] ^2+ + CO3^2- -> FeCO3 + 6H2O
  • Green precipitate
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10
Q

A 50.0 cm^3 sample of solution X was added to 50 cm^3 of dilute sulfuric acid and made up to 250 cm^3 of solution in a volumetric flask.
A 25.0 cm^3 sample of this solution from the volumetric flask was titrated with a 0.0205 mol dm^-3 solution of KMnO4

At the end point of the reaction, the volume of KMnO4 solution added was 18.70 cm^3.

State the colour change that occurs at the end point of this titration and give a reason for the colour change.

A
  • Pale green changes to purple solution

- just after the end - point, MnO4^- is in excess

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11
Q

Write an equation for the reaction between iron( II ) ions and manganate( VII ) ions.
Use this equation and the information given to calculate the concentration of iron( II ) ions in the original solution X.

( Volume of KMnO4 = 18.7 in a 25 cm^3 volume )
( Concentration of KMnO4 = 0.0205 )
( Volume of Fe^2+ = 50 cm^3 )

A
  • MnO4^- + 8H+ + 5Fe^2+ -> Mn^2+ + 4H2O + 5Fe^3+
  • Mol of MnO4 = 18.7 x 10^-3 x 0.0205
    = 3.8335 x 10^-4
  • Mol of Fe^2+ = 5 x 3.8335 x 10^-4
    = 1.91675 x 10^-3 ( 5:1 ratio )
  • Mol of Fe^2+ in 250 cm^3 = 10 x 1.91675 x 10^-3 = 1.91675 x 10^-2
  • Concentration of Fe^2+ = ( 1.91675 x 10^-2 ) / 50 x 10^-3 = 0.38335 mol dm^-3

= 0.383 mol dm^-3

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12
Q

A sample of hydrated nickel sulfate ( NiSO4.xH2O ) with a mass of 2.287 g was heated to remove all water of crystallisation.
The solid remaining had a mass of 1.344 g.

Calculate the value of the integer x.
Show your working.

A
  • 2.287 - 1.344 = 0.943 ( Mass of H2O )
  • NiSO4:
  • Mol = 1.344 / 154.8 = 8.682170543 x 10^-3
  • H2O:
  • Mol = 0.943 / 18 = 0.052388888889
  • Empirical NiSO4:

( 8.682170543 x 10^-3 ) / ( 8.682170543 x 10^-3 ) = 1

  • Empirical H2O:

( 0.05238888889 ) / ( 8.682170543 x 10^-3 )
= 6.034077381
= 6

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13
Q

Suggest how a student doing this experiment could check that all the water had been removed.

A
  • Reheat the experiment

- Check that the mass is unchanged

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14
Q

Propane-1,2-diol has the structure CH2( OH )CH( OH )CH3.
It is used to make polyesters and is one of the main substances in electronic cigarettes ( E-cigarettes ).
A sample of propane-1,2-diol was refluxed with a large excess of potassium dichromate( VI ) and sulfuric acid.

Draw the skeletal formula of propane-1,2-diol.

A
  • 2 horizontal diagonal lines to represent the carbon chain

- 2 vertical lines on the first and second vertices of the first line containing the OH molecules

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15
Q

Write an equation for this oxidation reaction of propane-1,2-diol under reflux, using [O] to represent the oxidizing agent.
Show the displayed formula of the organic product.

A
  • CH2OHCHOHCH3 + 3 [ O ] -> COOHCOCH3 + 2H2O

- ALL IN DISPLAYED FORMULA

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16
Q

Draw a labelled diagram to show how you would set up apparatus for refluxing.

A
  • Flask with a condenser on its neck
  • the flask is open
  • labelled; water in ( from the lower opening ) and water out ( on the top opening )
  • Heat on the bottom of the flask with an arrow pointing towards it
  • label the flask, condenser, water in and water out
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17
Q

Anti-bumping granules are placed in the flask when refluxing.
Suggest why these granules prevent bumping.

A
  • They help form smaller bubbles
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18
Q

Draw the structure of a different organic product formed when the acidified potassium dichromate(VI) is not in excess.

A
  • COCHOHCH3
  • DISPLAYED
  • THERE IS NO POSITIVE CHARGE ON THE FIRST CARBON
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19
Q

In order to obtain a pH curve, you are provided with a conical flask containing 25.0 cm^3 of a 0.100 mol dm^-3 carboxylic acid solution and a burette filled with 0.100 mol dm^-3 sodium hydroxide solution.
You are also provided with a calibrated pH meter.

State why calibrating a pH meter just before it is used improves the accuracy of the pH measurement.

A
  • Over time, meter doesn’t give accurate readings
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20
Q

Describe how you would obtain the pH curve for the titration.

( Of a carboxylic acid )

A
  • Measure the pH of the carboxylic acid
  • Add the alkali in known small portions
  • Stir the mixture
  • Measure the pH after every addition
  • Repeat this until alkali is in excess
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21
Q

A laboratory technician discovered four badly−labelled bottles, each containing one pure white solid.
Each bottle contained a compound of a different Group 2 metal ( magnesium, calcium, strontium and barium ).
Some tests were carried out on the solids or, if the compound was soluble, on the aqueous solution.
The results are given in the table.

Test

1 ) Added water

2 ) Solution or solid added to HCl ( aq )

3 ) Solution or solid added to NaOH ( aq )

4 ) Solution or solid added to H2SO4 ( aq )

Compound 1

1 ) Dissolves

2 ) Solution remains colourless

3 ) Solution gives a white precipitate

4 ) Solution has no visible change

Compound 2

1 ) Insoluble

2 ) Gives off carbon dioxide and a colourless solution forms

3 ) Solid remains insoluble

4 ) Gives off carbon dioxide gas and a white solid remains

Compound 3

1 ) Dissolves

2 ) Solution remains colourless

3 ) Solution gives a slight white precipitate

4 ) Solution slowly forms a slight white precipitate

Compound 4

1 ) Dissolves

2 ) Solution remains colourless and heat released

3 ) Solution has no visible change

4 ) Solution forms a white precipitate

One of the bottles has a very faint label that could be read as ‘ Magnesium Sulfate ’.
Use the information in the table to deduce which one of the four compounds is magnesium sulfate and explain your answer.

A
  • Compound 1

Explanation:

  • No visible change with H2SO4
  • Gives white precipitate with NaOH
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22
Q

The bottle containing Compound 2 has a ‘ TOXIC ’ hazard symbol.
Use the information in the table to identify Compound 2.
Explain both observations in the reaction with H2SO4( aq ).

( Test

1 ) Added water

2 ) Solution or solid added to HCl ( aq )

3 ) Solution or solid added to NaOH ( aq )

4 ) Solution or solid added to H2SO4 ( aq )

Compound 1

1 ) Dissolves

2 ) Solution remains colourless

3 ) Solution gives a white precipitate

4 ) Solution has no visible change

Compound 2

1 ) Insoluble

2 ) Gives off carbon dioxide and a colourless solution forms

3 ) Solid remains insoluble

4 ) Gives off carbon dioxide gas and a white solid remains

Compound 3

1 ) Dissolves

2 ) Solution remains colourless

3 ) Solution gives a slight white precipitate

4 ) Solution slowly forms a slight white precipitate

Compound 4

1 ) Dissolves

2 ) Solution remains colourless and heat released

3 ) Solution has no visible change

4 ) Solution forms a white precipitate )

A

Identity of Compound 2:

  • BaCO3

Explanation:

  • The carbonate ion releases CO2
  • But the BaSO4 formed is highly insoluble
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23
Q

Identify the compound that is strontium hydroxide.
Give an equation for the reaction of strontium hydroxide with sulfuric acid.

( Test

1 ) Added water

2 ) Solution or solid added to HCl ( aq )

3 ) Solution or solid added to NaOH ( aq )

4 ) Solution or solid added to H2SO4 ( aq )

Compound 1

1 ) Dissolves

2 ) Solution remains colourless

3 ) Solution gives a white precipitate

4 ) Solution has no visible change

Compound 2

1 ) Insoluble

2 ) Gives off carbon dioxide and a colourless solution forms

3 ) Solid remains insoluble

4 ) Gives off carbon dioxide gas and a white solid remains

Compound 3

1 ) Dissolves

2 ) Solution remains colourless

3 ) Solution gives a slight white precipitate

4 ) Solution slowly forms a slight white precipitate

Compound 4

1 ) Dissolves

2 ) Solution remains colourless and heat released

3 ) Solution has no visible change

4 ) Solution forms a white precipitate )

A
  • Compound 4

Equation:

  • Sr( OH )2 + H2SO4 - > SrSO4 + 2H2O
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24
Q

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.

( Diagram shows an apparatus of distillation with Alcohol + aqueous potassium dichromate ( VI ) vertically from the distillation apparatus looking like a burette which can drop the substance into the distillation apparatus, reagent P is the substance getting heated with a water bath of 60 celcius, a distillate is at the end of the apparatus which was obviously condensed )

Suggest the identity of reagent P.

A
  • H2SO4
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25
Q

State the chemical change that causes the solution in the flask to appear green at the end of the reaction.

( From a distillation process using alcohol and potassium dichromate )

A
  • Dichromate formed
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26
Q

Give one reason why using a water bath is better than direct heating with a Bunsen burner.

A
  • The alcohol is flammable
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27
Q

Suggest a reagent that could be used to confirm the presence of an aldehyde in the distillate.
State the observation you would expect to make if an aldehyde were present.

A

Reagent:

  • Tollens’

Observation:

  • Silver mirror
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28
Q

This question is about reactions of calcium compounds.

A pure solid is thought to be calcium hydroxide.
The solid can be identified from its relative formula mass.
The relative formula mass can be determined experimentally by reacting a measured mass of the pure solid with an excess of hydrochloric acid.
The equation for this reaction is:

Ca( OH )2 + 2HCl -> CaCl2 + 2H2O

The unreacted acid can then be determined by titration with a standard sodium hydroxide solution.
You are provided with 50.0 cm^3 of 0.200 mol dm^-3 hydrochloric acid.
Outline, giving brief practical details, how you would conduct an experiment to calculate accurately the relative formula mass of the solid using this method.

A
  • Mol HCl = 0.2 x 50 x 10^-3 = 0.01
  • Mol of Ca( OH )2 = 0.01 divided by 2 = 0.005
  • As acid is in excess, moles of Ca( OH )2 reacted will be less than this
  • Mass of Ca( OH )2 = 0.005 x 74.1 = 0.3705 g
  • Add 0.250 g ( Less than 0.371 ) of Ca( OH )2 to 50 cm^3 of 0.200 mol dm^-3 HCl in a conical flask
  • Titrate unreacted HCl with 0.200 mol dm^-3 NaOH until indicator changes colour
  • NaOH + HCl -> NaCl + H2O
  • moles of NaOH = moles of HCl
  • Moles of NaOH = 0.2 x 50 x 10^-3
  • = 0.01
  • Moles of Ca( OH )2 = 0.005 - 0.01
  • = - 0.005
  • Mr of Ca( OH )2 = 0.250 / -0.005
  • = - 50
  • = 50
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29
Q

A 3.56 g sample of calcium chloride was dissolved in water and reacted with an excess of sulfuric acid to form a precipitate of calcium sulfate.
The percentage yield of calcium sulfate was 83.4%.
Calculate the mass of calcium sulfate formed.
Give your answer to an appropriate number of significant figures.

A
  • Mol of CaCl2 = 3.56 / 111.1

= 0.03204320432

  • Mol of CaSO4 = 0.0320… x 83.4 x 10^-3

= 0.0267240324

  • Mass of CaSO4 = 0.0267… x 136.2

= 3.639813213

= 3.64 g

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30
Q

Calcium hydroxide is slightly soluble in water at room temperature.
As the temperature rises, the solubility decreases.
When the maximum amount of solid has dissolved at a particular temperature the solution is said to be saturated.
In an experiment, the solubility of calcium hydroxide was measured over a range of temperatures.
The results are shown in the graph.

( Graph shows an initial Solubility of calcium hydroxide of 0.18, from there, the graph decreases with a decreasing gradient, then decreases with a constant gradient and starts to level out as the graph ends )
( Y-axisis labelled “ Solubility of calcium hydroxide g / 100 cm^3 “ )
( X-axis is labelled “ temperature in celcius “ )

Use data from the graph to calculate the concentration, in mol dm^-3, of a saturated solution of calcium hydroxide at 30 °C.
Give your answer to 3 significant figures.
Show your working.

A
  • Draw a tangent at the point of 30 celcius

- Work out the gradient of the tangent at that point

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31
Q

You are given a sample of saturated calcium hydroxide solution.
Outline the practical steps that you would take to determine the solubility of calcium hydroxide in this solution.

A
  • Take a known volume of the saturated solution
  • Evaporate the filtrate to dryness
  • Weigh the residue
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32
Q

Calamine lotion can contain a mixture of zinc carbonate and zinc oxide in suspension in
water.
A manufacturer of calamine lotion claims that a sample contains 15.00 g of zinc carbonate and 5.00 g of zinc oxide made up to 100 cm3 with distilled water.

A chemist wanted to check the manufacturer’s claim.
The chemist took a 20.0 cm3 sample of the calamine lotion and added it to an excess of sulfuric acid.
The volume of carbon dioxide evolved was measured over time.
The chemist’s results are shown in the table.

Time / s

1 ) 0

2 ) 15

3 ) 30

4 ) 45

5 ) 60

6 ) 75

7 ) 90

8 ) 105

9 ) 120

10 ) 135

Volume / cm^3

1 ) 0

2 ) 125

3 ) 270

4 ) 380

5 ) 470

6 ) 530

7 ) 560

8 ) 570

9 ) 570

10 ) 570

Plot a graph of the results in the table.
The volume should be on the y-axis.
Draw a best-fit curve through all the points.

A
  • Sensible scales
  • All points plotted correctly
  • Smooth curve from 0 secs to at least 135 secs, line must past through or be close to all the points
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33
Q

Estimate the time taken for the reaction to be completed.

Calamine lotion question, a mixture of zinc carbonate and zinc oxide

A
  • Any value in the range of 91 and 105 secs

The time where the volume becomes constant

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34
Q

The volume of carbon dioxide in part (a) was measured at 293 K and at a pressure of 100
kPa.

Use information from your graph to calculate the maximum amount, in moles, of carbon dioxide evolved from the zinc carbonate in this 20.0 cm^3 sample.
The gas constant, R = 8.31 J K^-1 mol^-1

Show your working

( 570 is the maximum volume of CO2 produced )

A
  • PV = NRT
  • N = ( 100 x 10^3 ) ( 570 x 10^-6 ) / ( 8.31 ) ( 293 )
  • = 0.0234 mol
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35
Q

Use your answer to part ( i ) to calculate the mass of zinc carbonate in the 20.0 cm^3 sample of calamine lotion.
( If you were unable to complete part ( i ), you may assume that the amount of carbon dioxide evolved was 0.0225 mol.
This is not the correct answer. )

( Mol of CO2 = 0.02341025862 )

A
  • Mol of ZnCO3 = 0.02341025862
  • Mass = 0.0234… x 125.4 = 2.935646431

= 2.9 g

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36
Q

Calculate the difference between your answer to part ( ii ) and the manufacturer’s claim that there are 15.00 g of zinc carbonate in 100 cm^3 of the calamine lotion.
Express this difference as a percentage of the manufacturer’s claim.
( If you were unable to complete part ( ii ), you may assume that the mass of zinc carbonate in the
20 cm^3 sample of calamine lotion was 2.87 g.
This is not the correct answer. )

( Mass of ZnCO3 = 2.935646431 in 20 cm^3 )

A

Difference:

  • ( 15 divided by 5 ) - 2.93…
  • ( 20 cm^3 goes into 100 cm^3 five times )

= 0.06435356905

= 0.06

Percentage:

  • ( 0.0643…. / 3 ) x 100

= 2.1%

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37
Q

Draw a diagram of a suitable apparatus needed to perform the experiment outlined in part (a).
Include in your diagram a method for collecting and measuring the carbon dioxide.
The apparatus should be airtight.

( Reaction is ZnCO3 + H2SO4 )

A
  • Image shows a conical flask attached to a gas syringe with the bung on it
  • Conical flask contains a liquid that is labelled ZnCO3 + H2SO4
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38
Q

A peptide is hydrolysed to form a solution containing a mixture of amino acids.
This mixture is then analysed by silica gel thin-layer chromatography ( TLC ) using a toxic solvent.
The individual amino acids are identified from their Rf values.
Part of the practical procedure is given below.

  1. Wearing plastic gloves to hold a TLC plate, draw a pencil line 1.5 cm from the bottom of
    the plate.
  2. Use a capillary tube to apply a very small drop of the solution of amino acids to the mid-point
    of the pencil line.
  3. Allow the spot to dry completely.
  4. In the developing tank, add the developing solvent to a depth of not more than 1 cm.
  5. Place your TLC plate in the developing tank.
  6. Allow the developing solvent to rise up the plate to the top.
  7. Remove the plate and quickly mark the position of the solvent front with a pencil.
  8. Allow the plate to dry in a fume cupboard.

Parts of the procedure are in bold text.

( Wearing plastic gloves to hold a TLC plate, depth of not more than 1 cm, to the top, in a fume cupboard )

For each of these parts, consider whether it is essential and justify your answer.

A
  • ” Wearing plastic gloves “ is essential because it prevents contamination from the hands to the plate
  • ” Adding developing solvent to a depth of not more than 1 cm “ is essential, if the solvent is too deep, it will dissolve the mixture from the plate
  • ” Allowing the solvent to rise up the plate to the top “ isn’t essential as the RF value can be calculated if the solvent front doesn’t reach the top of the plate
  • ” Allowing the plate to dry in the fume cupboard “ is essential as the solvent is toxic
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39
Q

Outline the steps needed to locate the positions of the amino acids on the TLC plate and to determine their Rf values.

( A peptide is hydrolysed to form a solution containing a mixture of amino acids.
This mixture is then analysed by silica gel thin-layer chromatography ( TLC ) using a toxic solvent.
The individual amino acids are identified from their Rf values.
Part of the practical procedure is given below.

  1. Wearing plastic gloves to hold a TLC plate, draw a pencil line 1.5 cm from the bottom of
    the plate.
  2. Use a capillary tube to apply a very small drop of the solution of amino acids to the mid-point
    of the pencil line.
  3. Allow the spot to dry completely.
  4. In the developing tank, add the developing solvent to a depth of not more than 1 cm.
  5. Place your TLC plate in the developing tank.
  6. Allow the developing solvent to rise up the plate to the top.
  7. Remove the plate and quickly mark the position of the solvent front with a pencil.
  8. Allow the plate to dry in a fume cupboard. )
A
  • Spray the developing agent
  • Measure the distances from the initial pencil line to the spots
  • Measure the distance from the initial pencil line to the solvent front line
  • Rf value = x / y
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40
Q

Explain why different amino acids have different Rf values.

( A peptide is hydrolysed to form a solution containing a mixture of amino acids.
This mixture is then analysed by silica gel thin-layer chromatography ( TLC ) using a toxic solvent.
The individual amino acids are identified from their Rf values.
Part of the practical procedure is given below.

  1. Wearing plastic gloves to hold a TLC plate, draw a pencil line 1.5 cm from the bottom of
    the plate.
  2. Use a capillary tube to apply a very small drop of the solution of amino acids to the mid-point
    of the pencil line.
  3. Allow the spot to dry completely.
  4. In the developing tank, add the developing solvent to a depth of not more than 1 cm.
  5. Place your TLC plate in the developing tank.
  6. Allow the developing solvent to rise up the plate to the top.
  7. Remove the plate and quickly mark the position of the solvent front with a pencil.
  8. Allow the plate to dry in a fume cupboard. )
A
  • Amino acids have different polarities

- So have different solubility in the developing solvent

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41
Q

Iodide ions are present in seawater but at very low concentrations.
Certain types of seaweed are able to extract and concentrate iodide ions in their cells up to about 3% by mass.
This seaweed could be a source of the element iodine ( I2 ).
The following is an account of a method that was used to extract I2 from Laminaria seaweed.
The solvent tetrachloromethane ( CCl4 ) is no longer used because it is very toxic.

A. Collect and dry some seaweed.
Heat it very strongly in a crucible until a small quantity of ash is formed.
B. Add distilled or deionised water to the cooled ash in a beaker.
Boil the suspension for 5 minutes.
Allow to cool.
C. Filter the suspension.
D. Gradually add dilute sulfuric acid to the filtrate until the solution is acidic.
E. Add hydrogen peroxide solution.
A brown solution will be formed.
F. Transfer the solution to a separating funnel.
G. Add CCl4 to the funnel.
It forms a separate organic layer.
This organic layer changes from a colourless liquid to a purple solution.
H. Add the organic layer to an evaporating basin and place this in a fume cupboard.
Grey-black crystals of iodine will form as the solvent evaporates.

In Step A of the method, explain how it could have been confirmed that the decomposition of the seaweed was complete.

A
  • Re-heat the sample at least once

- The mass of ash shouldn’t change

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42
Q

The suspension was filtered in Step C.
Suggest an advantage of using vacuum filtration for this part of the method.

( Iodide ions are present in seawater but at very low concentrations.
Certain types of seaweed are able to extract and concentrate iodide ions in their cells up to about 3% by mass.
This seaweed could be a source of the element iodine ( I2 ).
The following is an account of a method that was used to extract I2 from Laminaria seaweed.
The solvent tetrachloromethane ( CCl4 ) is no longer used because it is very toxic.

A. Collect and dry some seaweed.
Heat it very strongly in a crucible until a small quantity of ash is formed.
B. Add distilled or deionised water to the cooled ash in a beaker.
Boil the suspension for 5 minutes.
Allow to cool.
C. Filter the suspension.
D. Gradually add dilute sulfuric acid to the filtrate until the solution is acidic.
E. Add hydrogen peroxide solution.
A brown solution will be formed.
F. Transfer the solution to a separating funnel.
G. Add CCl4 to the funnel.
It forms a separate organic layer.
This organic layer changes from a colourless liquid to a purple solution.
H. Add the organic layer to an evaporating basin and place this in a fume cupboard.
Grey-black crystals of iodine will form as the solvent evaporates. )

A
  • Saves time
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43
Q

Describe how, in Step D, the solution could have been tested to show when it became
acidic.
Your method should not contaminate the solution.

( Iodide ions are present in seawater but at very low concentrations.
Certain types of seaweed are able to extract and concentrate iodide ions in their cells up to about 3% by mass.
This seaweed could be a source of the element iodine ( I2 ).
The following is an account of a method that was used to extract I2 from Laminaria seaweed.
The solvent tetrachloromethane ( CCl4 ) is no longer used because it is very toxic.

A. Collect and dry some seaweed.
Heat it very strongly in a crucible until a small quantity of ash is formed.
B. Add distilled or deionised water to the cooled ash in a beaker.
Boil the suspension for 5 minutes.
Allow to cool.
C. Filter the suspension.
D. Gradually add dilute sulfuric acid to the filtrate until the solution is acidic.
E. Add hydrogen peroxide solution.
A brown solution will be formed.
F. Transfer the solution to a separating funnel.
G. Add CCl4 to the funnel.
It forms a separate organic layer.
This organic layer changes from a colourless liquid to a purple solution.
H. Add the organic layer to an evaporating basin and place this in a fume cupboard.
Grey-black crystals of iodine will form as the solvent evaporates. )

A
  • Place a pH probe into the solution

- Keep testing until pH drops below 7

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44
Q

Write an overall equation showing the oxidation of iodide ions to iodine in Step E using hydrogen peroxide solution in acidic conditions.

( No oxygen is evolved in this reaction. )

( Iodide ions are present in seawater but at very low concentrations.
Certain types of seaweed are able to extract and concentrate iodide ions in their cells up to about 3% by mass.
This seaweed could be a source of the element iodine ( I2 ).
The following is an account of a method that was used to extract I2 from Laminaria seaweed.
The solvent tetrachloromethane ( CCl4 ) is no longer used because it is very toxic.

A. Collect and dry some seaweed.
Heat it very strongly in a crucible until a small quantity of ash is formed.
B. Add distilled or deionised water to the cooled ash in a beaker.
Boil the suspension for 5 minutes.
Allow to cool.
C. Filter the suspension.
D. Gradually add dilute sulfuric acid to the filtrate until the solution is acidic.
E. Add hydrogen peroxide solution.
A brown solution will be formed.
F. Transfer the solution to a separating funnel.
G. Add CCl4 to the funnel.
It forms a separate organic layer.
This organic layer changes from a colourless liquid to a purple solution.
H. Add the organic layer to an evaporating basin and place this in a fume cupboard.
Grey-black crystals of iodine will form as the solvent evaporates. )

A
  • 2I^- + H2O2 + 2H^+ - > I2 + 2H2O
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45
Q

A diagram of a separating funnel is shown below.

( Diagram shows a separating funnel with an upper and lower layer of a solution, it also contains a tap on the bottom of it like a burette )

Aqueous solutions and tetrachloromethane are immiscible.
When added to the separating funnel they form two layers as shown in the diagram.
Prolonged shaking of the mixture allows iodine ( dissolved in the aqueous layer ) to pass into the tetrachloromethane layer.
The densities of some liquids are shown in the table.

Liquid

1 ) H2O

2 ) Filtrate from Step C

3 ) CCl4

4 ) Concentrated H2SO4

Density / gcm^-3

1 ) 1.0

2 ) 1.2

3 ) 1.6

4 ) 1.8

Explain briefly whether the upper layer or the lower layer in the separating funnel is likely to show a purple colouration after Step G.

( G. Add CCl4 to the funnel.
It forms a separate organic layer.
This organic layer changes from a colourless liquid to a purple solution. )

A
  • The lower layer as it will be the denser CCl4
46
Q

The iodine in the separating funnel quickly establishes an equilibrium after shaking the contents.
The concentrations of iodine in the aqueous layer and in the CCl4 layer become constant.
This is shown by the relationship

K = [ I2 in CCL4 ] / [ I2 in aqueous layer ]

A sample of Laminaria seaweed, mass 56.4 g, was processed by the method given.
After Step H, 1.673 g of iodine remained in the evaporating basin.
The volume of each layer was 50.0 cm3.
The seaweed contains 3.00% iodine by mass.
Calculate the value of K from these data. Show your working.

A
  • Iodine in seaweed = 56.4 x ( 3 / 100 ) = 1.692 g
  • Iodine left in the layer = 1.692 - 1.673 = 0.0190 g
  • K = ( 1.673 / ( 253.8 x 50 x 10^-3 )) / ( 0.0190 / ( 253.8 x 50 x 10^-3 )) = 88.1
    ( 253.8 = Mr of I2 )
47
Q

It has been suggested that cyclohexene could be used to extract the iodine from the aqueous layer.
Explain why this would not be a suitable solvent.

A
  • It would react with the iodine

- It is unsaturated

48
Q

The iodine produced at the end of Step H is impure and needs to be recrystallised from a suitable inert solvent.
Explain the essential feature to consider when choosing a suitable inert solvent for this recrystallisation.

A
  • Solubility is high when hot and low when cold

- So a significant quantity of crystals appear

49
Q

1,4-diaminobenzene is an important intermediate in the production of polymers such as Kevlar and also of polyurethanes, used in making foam seating.
A possible synthesis of 1,4-diaminobenzene from phenylamine is shown in the following figure.

( Image shows 4 steps, each with a benzene ring which has different functional groups on them, before step one it has a functional group of NH2, after step 1 , NHCOCH3, after step 2, NHCOCH3 and NO2, after step 3, NH2 and NO2, after step 4 , NH2 and NH2 )

A suitable reagent for step 1 is CH3COCl
Name and draw a mechanism for the reaction in step 1.

A
  • Mechanism name = nucleophilic addition

Mechanism:

  • Displayed formula of CH3COCl
  • An arrow from the bond between CO to the O
  • Displayed formula of C6H15NH2
  • NH2 contains a lone pair of electrons, the arrow from the lone pair goes to the carbon where the CO bond is present
  • The O from the CO bond gains a negative charge and has a lone pair of electrons, the lone pair of electrons move to the bond between CO, shown by an arrow, to reform the double bond of CO
  • Electrons from the CCl bond move to the Cl shown by an arrow, breaking off the Cl
  • The nitrogen has a positive charge when attached to the CH3COCl molecule, electrons from NH bond move to the N, shown by an arrow, breaking off the hydrogen
  • Show the final molecule in a displayed formula
  • CH3CONHC6H15
50
Q

The product of step 1 was purified by recrystallisation as follows.
The crude product was dissolved in the minimum quantity of hot water and the hot solution was filtered through a hot filter funnel into a conical flask.
This filtration removed any insoluble impurities.
The flask was left to cool to room temperature.
The crystals formed were filtered off using a Buchner funnel and a clean cork was used to compress the crystals in the funnel.
A little cold water was then poured through the crystals.
After a few minutes, the crystals were removed from the funnel and weighed.
A small sample was then used to find the melting point.
Give reasons for each of the following practical steps.

1 ) The minimum quantity of hot water was used

2 ) The flask was cooled to room temperature before the crystals were filtered off

3 ) The crystals were compressed in the funnel

4 ) A little cold water was poured through the crystals

A

1 ) To ensure the hot solution would be saturated

2 ) The yield is lower if warm

3 ) Air passes through the sample not just around it

4 ) To wash away soluble impurities

51
Q

The melting point of the sample in part ( b ) was found to be slightly lower than a data-book value.
Suggest the most likely impurity to have caused this low value and an improvement to the method so that a more accurate value for the melting point would be obtained.

( The product of step 1 was purified by recrystallisation as follows.
The crude product was dissolved in the minimum quantity of hot water and the hot solution was filtered through a hot filter funnel into a conical flask.
This filtration removed any insoluble impurities.
The flask was left to cool to room temperature.
The crystals formed were filtered off using a Buchner funnel and a clean cork was used to compress the crystals in the funnel.
A little cold water was then poured through the crystals.
After a few minutes, the crystals were removed from the funnel and weighed.
A small sample was then used to find the melting point.
Give reasons for each of the following practical steps.

1 ) The minimum quantity of hot water was used

2 ) The flask was cooled to room temperature before the crystals were filtered off

3 ) The crystals were compressed in the funnel

4 ) A little cold water was poured through the crystals )

A
  • Water

- Press the sample crystals between the filter papers

52
Q

The figure above is repeated here to help you answer the following questions.

( Image shows 4 steps )

( Phenylamine - > NHCOCH3benzene - > NHCOCH3benzeneNO2 - > NH2benzeneNO2 - > NH2benzeneNH2 )

( Each arrow is labelled “ Step 1, 2, 3 and 4 “ correspondingly )

In an experiment starting with 5.05 g of phenylamine, 4.82 g of purified product were obtained in step 1.
Calculate the percentage yield in this reaction.
Give your answer to the appropriate number of significant figures.

A
  • Mr of product = 135
  • Theoretical mass = 5.05 x ( 135 / 93 ) = 7.33
  • % yield = 4.82 / 7.33 x 100 = 65.8%
53
Q

A reagent for step 2 is a mixture of concentrated nitric acid and concentrated sulfuric acid, which react together to form a reactive intermediate.
Write an equation for the reaction of this intermediate in step 2.

( The figure above is repeated here to help you answer the following questions.

( Image shows 4 steps )

( Phenylamine - > NHCOCH3benzene - > NHCOCH3benzeneNO2 - > NH2benzeneNO2 - > NH2benzeneNH2 )

( Each arrow is labelled “ Step 1, 2, 3 and 4 “ correspondingly ) )

A
  • C6H15NHCOCH3 + NO2^+ -> C6H14( NHCOCH3 )NO2 + H^+
54
Q

Name a mechanism for the reaction in step 2.

C6H15NHCOCH3 + NO2^+ -> C6H14( NHCOCH3 )NO2 + H^+

A
  • Electrophilic substitution
55
Q

Suggest the type of reaction occurring in step 3.

( The figure above is repeated here to help you answer the following questions.

( Image shows 4 steps )

( Phenylamine - > NHCOCH3benzene - > NHCOCH3benzeneNO2 - > NH2benzeneNO2 - > NH2benzeneNH2 )

( Each arrow is labelled “ Step 1, 2, 3 and 4 “ correspondingly ) )

A
  • Hydrolysis
56
Q

Identify the reagents used in step 4.

( The figure above is repeated here to help you answer the following questions.

( Image shows 4 steps )

( Phenylamine - > NHCOCH3benzene - > NHCOCH3benzeneNO2 - > NH2benzeneNO2 - > NH2benzeneNH2 )

( Each arrow is labelled “ Step 1, 2, 3 and 4 “ correspondingly ) )

A
  • HCl
57
Q

During the preparation of aspirin, it is necessary to filter the crude product under reduced pressure.
Draw a diagram to show the apparatus you would use to filter the crude product under reduced pressure.
( Do not include the vacuum pump. )

A
  • Conical flask with a “ Side arm “ for the rubber tube

- Has a filter with a liquid soaking paper and a filter paper

58
Q

You are provided with a small sample of pure aspirin in a melting point tube.
Describe briefly how you would determine an accurate value for the melting point of aspirin.

A
  • Heat the melting point tube in an oil bath

- Slowly near the melting point

59
Q

N-phenylethanamide is used as an inhibitor in hydrogen peroxide decomposition and also in the production of dyes.
N-phenylethanamide can be produced in a laboratory by the reaction between phenylammonium sulfate and an excess of ethanoic anhydride:

A student carried out this preparation using 1.15 g of phenylammonium sulfate ( Mr = 284.1 ) and excess ethanoic anhydride.

( C6H5NH3 )2 SO4 + 2( CH3CO )2O - > 2C6H5NHCOCH3 + 2CH3COOH + H2SO4

Calculate the maximum theoretical yield of N-phenylethanamide that could be produced in the reaction.
Record your answer to an appropriate precision.
Show your working.

A
  • Mr of N - phenylethanamide = 135
  • Theoretical yield = 135 x 2 ( 1.15 / 284.1 )
    ( Multiplied by 2 as it contains 2 moles of that product, the Mr is multiplied by moles ( Mass / Mr )

= 1.09 g

60
Q

In the preparation, the student produced 0.89 g of N−phenylethanamide.
Calculate the percentage yield for the reaction.

( Theoretical yield = 1.09 )

A
  • ( 0.89 / 1.09 ) x 100

= 81.4%

61
Q

The student purified the crude solid product, N−phenylethanamide, by recrystallisation.
Outline the method that the student should use for this recrystallisation.

A
  • Dissolve the product in a minimum amount of volume of water
  • Hot water
  • Allow the solution to cool and for crystals to form
  • Filter off the pure product under reduced pressure
62
Q

Outline how you would carry out a simple laboratory process to show that the recrystallised product is a pure sample of N-phenylethanamide.

( Aspirin RP )

A
  • Measure the melting point
  • Use of melting point apparatus
  • Sharp melting point
63
Q

Assume that the reaction goes to completion.
Suggest two practical reasons why the percentage yield for this reaction may not be 100%.

( N-phenylethanamide is used as an inhibitor in hydrogen peroxide decomposition and also in the production of dyes.
N-phenylethanamide can be produced in a laboratory by the reaction between phenylammonium sulfate and an excess of ethanoic anhydride:

A student carried out this preparation using 1.15 g of phenylammonium sulfate ( Mr = 284.1 ) and excess ethanoic anhydride.

( C6H5NH3 )2 SO4 + 2( CH3CO )2O - > 2C6H5NHCOCH3 + 2CH3COOH + H2SO4 )

( Reference to aspirin practical )

A
  • Sample was still wet

- Some sample was list during recrystallisation

64
Q

The reaction to form N−phenylethanamide would happen much more quickly if the student used ethanoyl chloride instead of ethanoic anhydride.
Explain why the student might prefer to use ethanoic anhydride, even though it has a slower rate of reaction.

A
  • Violent reaction

- HCl gas isn’t released when ethanoic anhydride is used

65
Q

Salicylic acid can be used to make aspirin. Before using a sample of salicylic acid to
make aspirin, a student purified the acid by recrystallisation.
The method for recrystallisation is outlined below.

Step 1: The sample is dissolved in a minimum volume of hot water.

Step 2: The solution is filtered hot.

Step 3: The filtrate is cooled in ice to form crystals.

Step 4: The crystals are collected by filtration, washed with cold water and left to dry.

Explain the purpose of each underlined point.

( Underlined points :

  • Minimum volume
  • Hot water
  • Filtered hot
  • Cooled in ice )
A
  • Minimum volume, to obtain a saturated solution
  • Hot water, to increase yield
  • Filtered hot, to remove insoluble impurities
  • Cooled in ice, to increase the amount of crystals formed
66
Q

An engineer was trying to develop a new fuel for a motorboat by blending mixtures of different alcohols in order to find out which mixture released the most energy when used in the engine.
The engineer had a number of alcohols in unlabelled bottles. It was decided to identify the alcohols by determining their enthalpies of combustion and comparing these values with those from a data book.

Outline a simple practical experiment that the engineer could use, including the measurements to be taken, in order to determine the enthalpy of combustion for one of the unknown alcohols.
You do not need to include details of any calculations.

A
  • Weigh the spirit burner before and after combustion
  • Have water in a beaker
  • Measure the volume of water
  • Burn the alcohol to heat the water
  • Measure the temperature rise in water
67
Q

Other than heat loss to the surroundings, identify two major sources of error in the experiment.
Do not refer to the precision of the equipment.

( - Weigh the spirit burner before and after combustion

  • Have water in a beaker
  • Measure the volume of water
  • Burn the alcohol to heat the water
  • Measure the temperature rise in water )
A
  • Incomplete combustion

- Inadequate stirring

68
Q

The engineer found that the experimental values for the enthalpies of combustion of butan-1-ol and methylpropan-2-ol were very similar and so these values could not be used to distinguish between the two alcohols.
Identify a reagent that the engineer could use to distinguish between these two alcohols.
Give the observation in each case.

A
  • Reagent, acidified potassium dichromate
  • Butan - 1 - ol, turns from orange to green
  • ( Primary alcohol )
  • Methylpropan - 2 - ol, Remains orange
  • ( Tertiary alcohol )
69
Q

The filter in the air intake for the engine in the motorboat may become partially blocked by dust and debris.
Explain with the aid of an equation why combustion of methylpropan-2-ol under these circumstances would be of economic and environmental concern to the engineer.

A
  • C4H9OH + 4O2 -> 4CO + 5H2O
  • Engine wouldn’t run as efficiently
  • CO is a toxic gas
70
Q

Sodium bromate( V ) is a primary standard.
This means that its solution can be used to check the concentration of other solutions.
The half-equations for the reaction between bromate(V) ions and thiosulfate ions in the presence of acid are

2S2O3^2- -> S4O6^2- + 2e^-
BrO3^- + 6H^+ + 6e^- -> Br^- + 3H2O

Use these half-equations to deduce an overall equation for this reaction.

A
  • BrO3^- + 6H^+ + 6S2O3^2- -> 3S4O6^2-
    + Br^- + 3H2O

( Equal the amount of electrons in each equation so that they cancel out )

71
Q

A laboratory technician decided to use a 5.00 × 10^-3 mol dm^-3 solution of sodium bromate( V ) to check the concentration of a sodium thiosulfate solution that was labelled as 1.00 × 10^-3 mol dm^-3.
The sodium bromate( V ) solution was placed in the burette and 25.0 cm^3 of the sodium thiosulfate solution was pipetted into a conical flask.
Use the concentration of the sodium thiosulfate solution to calculate the expected titre value in this experiment.
Show your working.

A
  • Mol of sodium thiosulphate = 25 x 10^-3 x 1 x 10^-3 = 2.5 x 10^-5
  • Mol of sodium bromate = 1 / 6 x 2.5 x 10^-5 = 4.166666667 x 10^-6
  • Vol of sodium bromate = ( 4.166666667 x 10^-6 / 5 x 10^-3 ) x 1000

= 0.8333333334

  • Vol = 0.83 cm^3
72
Q

Consider the titre value that you have calculated in part ( b ).
Suggest one change to the experimental procedure in part ( b ) that would enable you to calculate a more accurate value for the concentration of the sodium thiosulfate solution.

( A laboratory technician decided to use a 5.00 × 10^-3 mol dm^-3 solution of sodium bromate( V ) to check the concentration of a sodium thiosulfate solution that was labelled as 1.00 × 10^-3 mol dm^-3.
The sodium bromate( V ) solution was placed in the burette and 25.0 cm^3 of the sodium thiosulfate solution was pipetted into a conical flask. )

( Titre value = 0.83 cm^3 )

A
  • Use a more dilute solution of sodium bromate
73
Q

Silver cyanide can be precipitated from sodium cyanide solution by adding an excess of silver nitrate solution.
Describe how you would obtain a pure dry sample of silver cyanide from this mixture.

A
  • Filter
  • Wash with water
  • Dry by pressing between filter paper or in air
74
Q

This question is about a toxic chloroalkane, X, that has a boiling point of 40 °C.
A student carried out an experiment to determine the Mr of X by injecting a sample of X from a hypodermic syringe into a gas syringe in an oven at 97 °C and 100 kPa.
The student’s results are set out in Table 1 and Table 2.

Table 1

Mass of hypodermic syringe filled with X before injection / g = 10.340

Mass of hypodermic syringe with left over X after injection / g = 10.070

Mass of X injected / g = to be calculated

Table 2

Volume reading on gas syringe before injection of X / cm^3 = 0.0

Volume of X in gas syringe after injection of X / cm^3 = 105.0

Volume of X / cm^3 = to be calculated

Complete Table 1 and Table 2 by calculating the mass and volume of X.

A
  • Mass of X injected / g = 10.340 - 10.070

= 0.270

  • Volume of X / cm^3 = 105.0
75
Q

X is known to be one of the following chloroalkanes: CCl4, CHCl3, CH2Cl2 or CH3Cl.
Justify this statement by calculating a value for the Mr of X and use your answer to suggest the most likely identity of X from this list.
Give your answer for the Mr of X to an appropriate precision.

(The gas constant R = 8.31 J K-1mol-1)

Mr of X:

( Mass of X = 0.270 )

Identity of X:

A
  • PV = nRT
  • n = PV / RT
  • n = ( 100 x 10^3 ) ( 105 x 10^-6 ) / ( 8.31 ) ( 97 + 273 )
  • n = 3.414967314 x 10^-3
  • Mr = 0.270 / 3.414967314 x 10^-3

= 79.06371428

= 79.1

Identity of X:

  • CH2Cl2
76
Q

Suggest a reason, other than apparatus inaccuracy, why the Mr value determined from the experimental results differs from the actual Mr.
Explain your answer.

( X is known to be one of the following chloroalkanes: CCl4, CHCl3, CH2Cl2 or CH3Cl.
Justify this statement by calculating a value for the Mr of X and use your answer to suggest the most likely identity of X from this list.
Give your answer for the Mr of X to an appropriate precision. )

A
  • The volume of the gas in the syringe is greater than the true volume
  • Mr = Mass / Mol = Mass x RT / PV, so if volume is too large, Mr will be too small
77
Q

Suggest, with a reason, an appropriate safety precaution that the student should take when using the toxic chloroalkane, X, in the experiment.

( X = CH2Cl2 )

A

Safety precaution:

  • Carry out in a fume cupboard

Reason:

  • To avoid toxic vapour
78
Q

A sample of strontium ore is known to contain strontium oxide, strontium carbonate and some inert impurities.
To determine the mass of strontium carbonate present, a student weighed a sample of the solid ore and then heated it in a crucible for 5 minutes.
The sample was allowed to cool 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

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.

A
  • CO2 gas escapes
79
Q

Use the data in the table to calculate the mass of strontium carbonate in the original ore sample.
Give your answer to an appropriate precision.

( 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 )

A
  • Mass of CO2 = 16.11 - 14.58 = 1.53 g
  • ( Mass of crucible and ore sample / g - Mass of crucible and sample after second heating / g ( When completely dried ) )
  • Mol of CO2 = 1.53 / 44.0 = 3.48 x 10^-2
  • Mol of SrCO3 = 3.48 x 10^-2
  • Mass of SrCO3 = 3.48 x 10^-2 x 147.6 = 5.13 g
80
Q

Each balance reading has an uncertainty of ±5.00 mg.
Calculate the percentage error in the initial mass of ore used.

  • ( Mass of SrCO3 = 5.13 )
  • ( Mass of CO2 = 1.53 )
A
  • 0.01 / ( 5.13 + 1.53 ) x 100 = 0.160 %
  • ( 5.00 mg = 0.005 g )
  • ( 0.005 x 2 = 0.01 ) ( Two readings are required from a balance )
  • ( 5.13 + 1.53 = The total mass of SrCO3 and CO2 )
81
Q

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 your answer.

  • ( SrCO3 ⟶ SrO + CO2 )
  • ( Mass of SrCO3 = 5.13 )
  • ( Mass of CO2 = 1.53 )
A
  • ( 6.26 = ( 5.13 + 1.53 ) )
  • ( 346 mg = 0.346 g )
  • 6.26 - 0.347 - 5.13 = 0.783
  • Mass of SrO = 0.783 g
  • All SrCO3 reacted because it was heated to a constant mass ( Assumption made during calculation )
82
Q

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.

A
  • 3SrO + 2Al - > Al2O3 + 3Sr ( Equation )
  • Aluminium acts as a reducing agent ( Role of Al in this equation )
  • Sr is collected as a vapour because
  • Al2O3 is an ionic lattice, so has strong ionic attractions
  • than Sr which is a metallic structure with weaker bonding
83
Q

Ammonium chloride, when dissolved in water, can act as a weak acid as shown by the following equation.

NH4^+( aq ) - > NH3( aq ) + H^+( aq )

The following figure shows a graph of data obtained by a student when a solution of sodium hydroxide was added to a solution of ammonium chloride.
The pH of the reaction mixture was measured initially and after each addition of the sodium hydroxide solution.

  • ( Graph shows “ pH “ against “ volume of NaOH added / cm^3 “ )
  • ( Line graph starts at 4.6 pH with 0 NaOH added, it increases as NaOH increases )
  • ( Gradient of graph decreases a bit but still increases from a 1 cm^3 to 9 cm^3 of NaOH added )
  • ( Gradient increases once more at 9.5 cm^3 of NaOH added to around 10.5 cm^3 of NaOH added )
  • ( Graph flattens from 10.5 cm^3 to 12 cm^3 of NaOH added )
  • ( At a pH of 13.2 )

Suggest a suitable piece of apparatus that could be used to measure out the sodium hydroxide solution.
Explain why this apparatus is more suitable than a pipette for this purpose.

A
  • Apparatus: Burette

- Because it can deliver variable volumes

84
Q

Use information from the curve in the figure above to explain why the end point of this reaction would be difficult to judge accurately using an indicator.

  • ( Graph shows “ pH “ against “ volume of NaOH added / cm^3 “ )
  • ( Line graph starts at 4.6 pH with 0 NaOH added, it increases as NaOH increases )
  • ( Gradient of graph decreases a bit but still increases from a 1 cm^3 to 9 cm^3 of NaOH added )
  • ( Gradient increases once more at 9.5 cm^3 of NaOH added to around 10.5 cm^3 of NaOH added )
  • ( Graph flattens from 10.5 cm^3 to 12 cm^3 of NaOH added )
  • ( At a pH of 13.2 )
A
  • The change in pH is gradual at the end point

- An indicator would change colours over a range of volumes of sodium hydroxide

85
Q

The pH at the end point of this reaction is 11.8.
Use this pH value and the ionic product of water, Kw = 1.0 × 10^-14 mol^2 dm^-6, to calculate the concentration of hydroxide ions at the end point of the reaction.

A
  • [ H^+ ] = 10^-11.8
  • = 1.58 x 10^-12
  • ( kw = [ H^+ ] [ OH^- ] )
  • [ OH^- ] = 1 x 10^-14 / 1.58 x 10^-12
  • = 6.33 x 10^-3 mol dm^-3
86
Q

The expression for the acid dissociation constant for aqueous ammonium ions is

Ka = [ NH3 ] [ H^+ ] / [ NH4^+ ]

The initial concentration of the ammonium chloride solution was 2.00 mol dm^-3.
.
Use the pH of this solution, before any sodium hydroxide had been added, to calculate a value for Ka.

  • ( Initial pH = 4.6 )
  • ( pH was obtained from the graph )
A
  • ( At this point [ NH3 ] = [ H^+ ] )
  • Ka = [ H^+ ]^2 / [ NH4^+ ]
  • [ H^+ ] = 10^-4.6
  • = 2.51 x 10^-5
  • Ka = [ 2.51 x 10^-5 ]^2 / 2
  • ( 2.00 is the initial concentration of ammonium chloride )
  • Ka = 3.15 x 10^-10 mol dm^-3
87
Q

A solution contains equal concentrations of ammonia and ammonium ions.
Use your value of Ka from part ( d ) to calculate the pH of this solution. Explain your working.

  • ( Ka = 3.15 x 10^-10 mol dm^-3 )
A
  • ( When [ NH3 ] = [ NH4^+ ], they cancel out )
  • ( Ka = [ NH3 ] [ H^+ ] / [ NH4^+ ] )
  • ( So Ka = [ H^+ ] )
  • pH = - log [ H^+ ]
  • pH = - log ( 3.15 x 10^-10 )
  • pH = 9.50
88
Q

A student was given unlabelled samples of pentan-1-ol, pent-1-ene, pentanoic acid and pentanal.
Name the reagent( s ) that the student could use to identify the sample that was pent-1-ene.
Describe the observation( s ) that the student would make to confirm this.

A
  • Bromine water

- Orange to colourless

89
Q

Name the reagent( s ) that the student could use to identify the sample that was pentanoic acid.
Describe the observation( s ) that the student would make to confirm this.

( A student was given unlabelled samples of pentan-1-ol, pent-1-ene, pentanoic acid and pentanal. )

A
  • Sodium hydrogencarbonate

- Effervescence

90
Q

Name the reagent( s ) that the student could use to identify the sample that was pentanal.
Describe the observation( s ) that the student would make to confirm this.

( A student was given unlabelled samples of pentan-1-ol, pent-1-ene, pentanoic acid and pentanal. )

A
  • Tollens’ reagent

- Silver mirror

91
Q

The student deduced that the spectrum in the image below was that of pentanal.

  • ( Spectrum shows a large peak at 1750 cm^-1 )

Justify this deduction and suggest why this spectrum cannot be that of pentan-1-ol, pentanoic acid or pent-1-ene.

A
  • Absorption at 1680 - 1750 cm^-1 caused by C = O
  • No absorption at 1620 - 1680 cm^-1 which is caused by C = C
  • No absorption at 3230 - 3550 cm^-1 which is cased by - OH alcohol
  • No absorption at 2500 - 3000 cm^-1 which is caused by - OH acid
92
Q

The iron( II ) ions in well-water can be removed by oxidation using potassium manganate( VII ) in alkaline solution.
A mixture containing solid iron( III ) hydroxide and solid manganese( IV ) oxide is formed.
These solid products can be removed by filtration under reduced pressure.
Draw a diagram of the apparatus used for this filtration.
Do not include the apparatus used to reduce the pressure.

A
  • Flask with side - arm

- Buchner funnel and horizontal filter paper

93
Q

An equation representing the oxidation reaction is given below.

3Fe^2+( aq ) + KMnO4( aq ) + 5OH^-( aq ) + 2H2O( l ) - > 3Fe( OH )3( S ) + MnO2( S ) + K^+( aq )

Calculate the mass, in grams, of KMnO4 required to react with the iron( II ) ions in 1.00 dm^3 of well-water that has an iron( II ) concentration of 0.225 mol dm^-3.
Give your answer to the appropriate precision.
Show your working.

A
  • Mr KMnO4 = 158

- Mass KMnO4 = 0.225 / 3 x 158 = 11.9 g

94
Q

In practice, a slight excess of potassium manganate( VII ) is used to treat the well-water.
Although this treated water is safe to drink, this excess of potassium manganate( VII ) is undesirable.
Suggest one reason, other than colour, why the excess is undesirable.

A
  • Unpleasant taste
95
Q

Suggest one reason why the colour of potassium manganate( VII ) solution can be a source of error when using a volumetric ( graduated ) flask to prepare a standard solution.

A
  • It will be difficult to see the meniscus
96
Q

This question is about the chemical properties of chlorine, sodium chloride and sodium bromide.
Sodium bromide reacts with concentrated sulfuric acid in a different way from sodium chloride.
Write an equation for this reaction of sodium bromide and explain why bromide ions react differently from chloride ions.

A
  • Equation: 2NaBr + 2H2SO4 - > Na2SO4 + Br2 + SO2 + 2H2O
  • Br^- ions are bigger than Cl^- ions
  • So can be more easily oxidised compared to Cl^-
97
Q

A colourless solution contains a mixture of sodium chloride and sodium bromide.
Using aqueous silver nitrate and any other reagents of your choice, develop a procedure to prepare a pure sample of silver bromide from this mixture.
Explain each step in the procedure and illustrate your explanations with equations, where appropriate.

A
  • Add silver nitrate to form precipitates
  • Ag^+ + Cl^- - > AgCl
  • White precipitate
  • Ag^+ + Br^- - > AgBr
  • Cream precipitate
  • Add excess dilute ammonia
  • Silver chloride precipitates dissolve
  • AgCl + 2NH3 - > Ag( NH3 )2^+ + Cl^-
  • Filter off AgBr precipitate
  • Wash and dry it
98
Q

Write an ionic equation for the reaction between chlorine and cold dilute sodium hydroxide solution.
Give the oxidation state of chlorine in each of the chlorine-containing ions formed.

A
  • Cl2 + 2HO^- - > OCl^- + Cl^- + H2O

Oxidation state of Cl:

  • OCl^- = + 1
  • Cl^- = - 1
99
Q

A sample of solid chromium( III ) hydroxide displays amphoteric character when treated separately with dilute hydrochloric acid and with dilute aqueous sodium hydroxide.
Write an ionic equation for each of these reactions.
Include the formula of each complex ion formed.
Describe the changes that you would observe in each reaction.

A
  • Cr( OH )3 + 3H2O + 3H^+ - > [ Cr( H2O )6 ]^3+
  • Green solid
  • Forms a green solution
  • Cr( OH )3 + 2H2O + OH^- - > [ Cr( H2O )2( OH )4 ]^-
  • Forms a green solution
100
Q

Aqueous solutions of copper( II ) sulfate and cobalt( II ) sulfate undergo ligand substitution reactions when treated separately with an excess of dilute aqueous ammonia.
Write equations for these reactions.
Include the formulae for any complex ions.
Describe the changes that you would observe in each reaction.

A
  • [ Cu( H2O )6 ]^2+ + 4NH3 - > [ Cu( H2O )2( NH3 )4 ]^2+ + 4H2O
  • Blue solution
  • Dark blue solution
  • [ Co( H2O )6 ]^2+ + 6NH3 - > [ Co( NH3 )6 ]^2+ + 6H2O
  • Pink solution
  • Brown solution
101
Q

A student calculated that a value for the enthalpy change of neutralisation is -51.2 kJ mol^-1.
The design of a possible hand-warmer using hydrochloric acid and sodium hydroxide was discussed.
It was proposed that 500 cm^3 of hydrochloric acid should be used in a flexible, sealed plastic container with a breakable tube of solid sodium hydroxide also in the container.
On breaking the tube, the sodium hydroxide would be released, react with the acid and produce heat.
A 40 °C temperature rise was thought to be suitable.

Calculate the heat energy, in J, required to raise the temperature of the reaction mixture by 40 °C.
Assume that the reaction mixture has a density of 1.00 g cm–3 and a specific heat capacity of 4.18 J K^-1 g^-1.
Assume that all of the heat energy given out is used to heat the
reaction mixture.

A
  • Q = M x S x delta T
  • Q = 500 x 4.18 x 40
  • Q = 83600 J
102
Q

Use your answer from part ( a ) and the value for the enthalpy change of neutralisation of - 51.2 kJ mol^-1 to calculate the minimum amount, in moles, and hence the minimum mass of sodium hydroxide required in the breakable tube.
Show your working.

( Q = 83600 J )

A
  • Delta H = Q / moles of NaOH
  • Moles of NaOH = Q / Delta H
  • Moles of NaOH = 83600 x 10^-3 / 51.2 ( J - > KJ )
  • Moles of NaOH = 1.63
  • Mass of NaOH = 1.63 x 40 = 65.2 g
103
Q

Use the amount, in moles, of sodium hydroxide from part ( b ) to calculate the minimum concentration, in mol dm^-3, of hydrochloric acid required in the 500 cm^3 of solution used in the sealed container.

( Moles of NaOH = 1.63 )

A
  • Concentration of HCl = 1.63 / 500 x 10^-3

- = 3.26 mol dm^-3

104
Q

Suggest one possible risk to a person who uses a hand-warmer containing sodium hydroxide and hydrochloric acid.

A
  • Container splitting and releasing corrosive chemicals
105
Q

A commercial hand-warmer uses powdered iron sealed in a plastic container.
A valve allows air to enter the container, and oxygen in the air reacts slowly with the iron to form solid iron( lll ) oxide.
The heat released warms the container.

Write an equation for this reaction between iron and oxygen to form iron( lll ) oxide.

A
  • 4Fe + 3O2 - > 2Fe2O3
106
Q

One version of an iron-oxygen hand-warmer advertises that it is designed to stay warm for up to four hours.
Other than by increasing the amount of iron in the container, state one change to the iron in the hand-warmer that would increase this time.
Explain why this change to the iron might not be an advantage.

A
  • Iron powder particle size can be increased ( Would increase the time of the hand-warmer )
  • So a slower release of energy
107
Q

Another type of hand-warmer uses sodium thiosulfate.
Sodium thiosulfate is very soluble in water at 80 °C but is much less soluble at room temperature.
When a hot, concentrated solution of sodium thiosulfate is cooled it does not immediately crystallise.
The sodium thiosulfate stays dissolved as a stable ’super-saturated’ solution until crystallisation is triggered.
Heat energy is then released when the sodium thiosulfate crystallises.
This type of hand-warmer is re-usable.
Suggest one environmental advantage that a sodium thiosulfate hand-warmer has over the other two types.

A
  • Conserves resources
108
Q

Describe the two steps that you would take to make the sodium thiosulfate hand-warmer ready for re-use.

A
  • Step 1: Heat to 80 degrees celcius

- Step 2: Allow it to cool before using it again

109
Q

One cell that has been used to provide electrical energy is the Daniell cell.
This cell uses copper and zinc.
The conventional representation for the Daniell cell is:

Zn( s ) | Zn^2+( aq ) | | Cu^2+( aq ) | Cu( s )

The e.m.f. of this cell under standard conditions is +1.10 V.
Deduce the half-equations for the reactions occurring at the electrodes.

A
  • At Zn electrode: Zn( s ) - > Zn^2+( aq ) + 2e^-

- At Cu electrode: Cu^2+( aq ) + 2e^- - > Cu( s )

110
Q

A Daniell cell was set up using 100 cm^3 of a 1.0 mol dm^-3 copper( II ) sulfate solution.
The cell was allowed to produce electricity until the concentration of the copper( II ) ions had decreased to 0.50 mol dm^-3.
Calculate the decrease in mass of the zinc electrode.
Show your working.

A
  • Moles of copper = 100 x 10^-3 x 0.5 = 0.05
  • Moles of zinc = 0.05
  • Mass of zinc lost = 0.05 x 65.4 = 3.27 g
111
Q

You are provided with the Daniell cell referred to in part ( b ), including a zinc electrode of known mass.
Briefly outline how you would carry out an experiment to confirm your answer to part ( b ).

( Concentration of copper ions = 0.5 )

A
  • Allow cell to discharge until [ Cu^2+ ] is 0.5
  • Confirmed by colorimetric measurement or other suitable method
  • ( Solution colourless )
  • Weigh the Zn electrode before and after the experiment

Chemistry (13 decks)

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