TRANSNZOIA WEST DISTRICT CHEMISTRY PRACTICAL QUESTIONS AND ANSWERS

TRANSNZOIA WEST DISTRICT CHEMISTRY PRACTICAL QUESTIONS

CONFIDENTIAL

INSTRUCTIONS

ACCESS TO

• 1M NaOH
• 1M NH₄OH
• 1M HCL
• 0.01m PB (NO₃)₂
• Source of heat
• PH chart (PH=1 to 14)
• 10ml of solution K
• Sodium hydrogen carbonate

Question 1.

1. Solution J 100cm³
   
2. Burette
   
3. Solution K100cm³
   
4. Pipette
   
5. 2 conical flasks
   
6. Filter funnel
   
7. Retort stand
   

PREPARATION OF SOLUTIONS

1. Solution J – Dissolve 17g of ammonium iron (ii) sulphate in 50cm³ of 2M H₂SO₄ dilute to 1dm³

2. Solution K-KMnO_{4 –} Dissolve 1.6g of potassium manganate vii in 20cm³ of 2 MH₂SO₄ dilute to 1dm³

3. Solution R – Dissolve 40g of sodium thiosulphate in 1dm³ of solution

4. Solution S – Dissolve 172cm³ of concentrated hydrochloric acid in 1dm³ of solution

5. Solid Y is aluminium sulphate

6. Solid Z is oxalic acid.

1. You are provided with:

• Solution M₁ aqueous solution of a monobasic acid, HB containing 1.62425, of the acid dissolve in 250cm³ of the solution
• 0.208M sodium hydroxide solution.

You are required to determine

a) The molarity of the acid

 b) The RFM of the acid and the RAM of B in HB (H=1, C=12, O=16)

Procedure

Pipette 25cm³ of solution M1 into a clean dry conical flask. Add 2 drops of phenolphthalein indicators. Fill the burette with solution Q and titrate against solution M₁

Repeat the procedure two more times and complete the table below:

a) Determine the average volume of solution Q used

 b) Write an equation for the reaction between solution M₁ and Q

 c) Calculate:

i) The number of moles of Q used

ii) The number of moles of M₁ used

iii) The molarity of solution M₁

 d) Determine;

i) The RFM of acid

ii)The RAM of element B

2. You are provided with:

• 2M hydrochloric acid, solution M₂
  

• Magnesium ribbon.

You are required to determine;

 i) The rate of the reaction between Hydrochloric acid and magnesium

ii) The mass of 2cm of magnesium ribbon

Procedure II

Using a clean measuring cylinder, measure 60cm³ of 2M hydrochloric acid, solution M₂ and

place it into a clean conical flask. Cut a 2cm piece of magnesium ribbon provided and place

into the conical flask containing 2M hydrochloric acid and immediately start the slop- watch.

Measure and record the time taken for the magnesium ribbon to completely react with the

hydrochloric acid in table II below. Repeat the procedure using 50, 40, 30 and 20cm3
portions

of 2M hydrochloric acid adding distilled water and complete the table below:

a) Table II

b) Plot a graph of ½ against volume of 2M hydrochloric acid used

c) From your graph determine the time taken for the ribbon to disappear when 36cm³ of 2M

hydrochloric acid were used

 d) In terms of rate of reaction, explain the shape of your graph

3. You are provided with solids. You are required to carry out the tests shown below and write

your observations and inference in the spaces provided. Identify any gases given out.

 a) Place a small amount of solid S in a dry test tube and heat strongly

b) Place a spatula end- full of S in a boiling tube. Add about 5cm³ of distilled water and shake.

Divide the resultant mixture into 4 portions

i) to the first portion, add nitric acid followed by Barium nitrate solution

ii) To the second portion, add nitric acid followed by lead (II) nitrate solution. Warm the mixture

iii) To the forth portion, add aqueous ammonia drop wise until excess

3. b) You are provided with solid F. Carry out the texts below. Write your observations and

inferences in the space provided.

Dissolve a spatula full of solid F in about 4cm³ of distilled water and divide it into three parts.

i) To 2cm³ of solution, add 5 drops of bromine water

ii) To the second portion add a spatula full of sodium hydrogen carbonate

TRANSNZOIA WEST DISTRICT CHEMISTRY PRACTICAL ANSWERS

Q1. i) Complete table with 3 titrations done – 1 mark

 ii) Incomplete table with 2 titrations done – ½ mark

 iii) Incomplete table with 1 titration done – 0 marks

 Penalties

1. Wrong arithmetic
   
2. Inverted table
   
3. Unrealistic values
   

Penalize ½ mark for each to maximum of ½ mark

 Decimals (1 mark)

Conditions

1. Accept either 1 or 2 decimal point constitently.
   
2. If 2 decimal point used the 2^nd decimal point can only be 0 or 5
   

Accuracy 1 mark

 Compare any litre values in the 3^rd row with the school value (sv)

Conditions

 i) If within I 0.1cm3 of S.V 1 mark

 ii) If within I 0.2 of S.V ½ mark

 iii) Beyond I 0.2 of SV 0 mark

N.B If there is wrong arithmetic in the table compare the SV with the correct value and credit accordingly

 d) Principle of averaging 1 mark

Values averaged must be shown and must be within I 0.2cm3 of each other

 Conditions

 i) 3 values averaged and consistent – 1 mark

 ii) 3 values done and only 2 possible averaged 1 mark

 iii) 2 titrations done and averaged 1 mark

 iv) 2 titrations done inconsistent ½ mark

 v) 3 titrations done and possible but only two averaged 0 mark

 e) Final answer 1mark

NB Compare the SV

i) If within I0.1 of SV 1 mark

ii) If within I 0.2 of SV ½ mark

If beyond I 0.2 of SV 0 mark

If the candidate has averaged wrong values, pick the correct value if any, average and

credit accordingly

B. HB_(aq) + NaOH_(aq) ___________ NaB_(aq) + H₂O_(L) 1 mark

C. i) 0.2075 X Volume = Moles 1 mark

1000

 ii) Reacting ratio 1: 1

჻ Moles of T = answer in C (i) above

 iii) Answer in b(ii) above X 1000

25

 d) i) 1.62425g __________ 250cm3

6.497g/l __________ 1000cm3

M = g/l

Mm

჻ mm = 6.497

 Answer in b(ii) above

ii) HB = answer in d(ii) – 1

B =

Question 2.

1. 120cm³ of solution R
   
2. 80cm³ of solutions
   
3. 250cm³ of tap water
   
4. 25 or 50ml measuring cylinder
   
5. 100cm³ glass beaker
   
6. 5 x5cm piece of white paper
7. Stop watch or clock
   

Q2. Table II

 a) Table

Marking areas

i) Complete table

Penalties

• Penalize ½ mark for each space not filled
  
• Reject fractions for ⁱ/_t and award a max of 1 ½ for table
  
• If fractions appear followed by an extra column of decimals, ignore the fractions and award accordingly
  
• Penalize ½ mark each for wrong arithmetic in the value of ⁱ/_t not within an error of +-2 units in the 3^rd decimal place unless it divides exactly
  
• Accept reciprocals given to at least 3 decimal places otherwise penalize ½ mark each for
  

rounding off to the 2^nd decimal place to a max of 1 mark unless it divides exactly

• Penalize ½ mark for every reading < 5 and > 120 seconds in the time row
  
• Penalize ½ mark for each entry not in seconds
  

 ii) Use of decimals

(Tied to the 4^th row only)

– Accept a whole numbers or decimals up to the 2^nd decimal place only used

consistently, otherwise penalize fully

 iii) Accuracy

(Tied up to the 4^th row only)

– Compare the candidates 1^st reading to the S.V and if within +- 2 sec, award 1

mark, otherwise penalize fully

 iv) Trend

(Tied to the 4^th row only)

• Award 1 mark if time is continuously increasing otherwise penalize fully
  

 b) Graph

 i) Labeling of both axes

 Condition

• Penalize ½ mark for wrong units used in any of the axis
  
• Penalize ½ mark for inverted axes
  
• Accept if units are not shown. Otherwise if shown they MUST be correct
  
• Both axes MUST be labeled
  

ii) Scale

• Area covered by the actual plots including the origin should be 2/3 more of the squares
  

provided in both axes

• The scale interval should be consistent
  

iii) Plotting

• Award 1 mark if 4 or 6 plots are correctly plotted
  
• Award ½ mark if 2 or 3 plots are correctly plotted
  
• Accept plots even if the axes are inverted
  
• Accept rounding off the values of i/t to the 3^rd decimal point when plotting
  

iv) Line

– Accept a straight line passing through at least 2 points correctly plotted and through

the origin (0,0) for 1 mark or if extrapolated can pass through the origin

 c) – Showing ⁱ/_t on the graph – Stating the correct reading of ⁱ/_t at 36cm³

– Applying the expression that time = ⁱ/_t correct reading – Correct answer

 d) Rate decrease with decrease in concentration of hydrochloric acid or vice versa

OR

Rate and concentration are directly proportional

Condition

• Tied to the correct graph or trend in the table
  
• If volume is used in place of conc. Award ½ mark
  

3. a)

b)

iii) To the third portion add a few drops of acidified potassium dichromate (VI)

Q 1. Table 1 (5 mks)

 a) Complete table (1 mk)

 – Penalize ½ mk for arithmetic error or unrealistic value to a maximum of ½ marks

 b) Use of decimal (1 mark)

 – Candidates to use 1 d.p or 2 d.p throughout in 1^st and 2^nd rows

 c) Accuracy (1 mark)

± 0.2 the S.V  ½ NB Any one value from the table

± 0.1 the S.V  1

 d) Principles of averaging (1 mark)

– I + II + III
 ½

3

– Correct answer  ½

 e) Final answer

Average of the candidate compared with school value (S.V)

± 0.2  ½

± 01  1

ii) Moles of N = 25 x 0.1
 ½

1000

= 0.0025  ½

iii) HCL _(aq) + NaOH _(aq) NaCL _(aq) + H₂O _(L)

Balanced  ½

State symbols ½

iv) HCL: NaOH  1

1 : 1

 Moles of M = 1×0.0025
 ½

1

= 0.0025  ½

v) Average titre 0.0025

1000 cm³ ?

= 1000×0.0025
 ½

Average titre

= Correct answer  ½

vi) Answer (V) x 36.5
 ½

1

= Correct answer  ½

Table II

1. As in table I
   

b) Answer in (v) x Titre
 ½

1000

= Correct answer  ½

 c) 2HCL _(aq) + Na₂CO_{3 (aq)} 2NaCL _(aq) + H₂O _(l) + CO_{2 (g)}

Balanced  ½

State symbol  ½

d) HCL: Na₂CO₃

2 : 1 1

1x Answer in (b)
 ½

2

= Correct answer ½

e) 1000 x Answer in (d)
 ½

25

= Correct answer ½

f) 14.3g/litre 1

g) R + M = Mass in g/h

Molarity

= 14.3
 ½

Answer in (e)

= Correct answer  ½

h) Answer in (g) = 106 + 18x  ½

18x = Answer in (g) – 106

x = Answer in (g) – 106
 ½

18

= Correct answer  1(should be a whole number)

Q 2. Table

 Each entry ½ mark

• Penalize ½ mark to a maximum of 1 mark for unrealistic values
  
• Penalize ½ mark mixing decimal numbers and whole numbers
  

a) i) Labeling ( ½ mark)

 ii) Scale (½ mark)

 iii) Plotting (2 marks)

 iv) Line/ curve (1 mark)

b) i) 1. x 5  ½

1000

= 0.005  ½

c) Pb²⁺ : I^–

0.0025: 0.005  1

1 : 2  1

d) Pb (NO₃)_2(aq) + 2KI _(aq) PbI _(s) + 2KNO_{3 (aq)}

Balanced  1

States symbol 1

e) Pb²⁺_(aq) + 2I^–(aq) PbI_2(s)

Balance  ½

States  ½

f) To make the setting of precipitate faster 1