KAKAMEGA CENTRAL DISTRICT CHEMISTRY PRACTICAL QUESTIONS AND ANSWERS

KAKAMEGA CENTRAL DISTRICT CHEMISTRY PRACTICAL QUESTIONS

CONFIDENTIAL

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
  

PREPARATION OF SOLUTIONS:  

1. Solution J

Dissolve 17g of ammonium iron (II) sulphate in 50cm³ of 2M H₂SO₄ dilute to 1dm³

2. Solution K KMnO₄

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.

Each candidate will require:

Q1.

1.  Solution J – 100cm³

2.   Burette

3.   Solution K- 100cm³

4.   Pipette

5.  2 conical flasks

6.  Filter funnel

7.  Retort stand

1.  You are provided with:

 Solution J:xM ammonium iron(II)sulphate solution

 Solution K: 0.02M potassium manganate (VII)solution

You are required to determine:

 -The molarity, x of the ammonium iron (II) sulphate

 – The amount of water of crystallisation, N in ammonium iron (II) sulphate

 -The formula mass of ammonium iron (II)sulphate.

Procedure

The ammonium iron (II) sulphate, (NH₄)₂SO₄FeSO₄nH₂O solution provided was made by

dissolving 8.5g of the salt in 50.0cm³ of dilute sulphuric(VI)acid, then making the solution

to 250cm³ using distilled water.

Fill the burette with solution K. Pipette 25cm³ of solution J and release into a conical flask.

Titrate J against K until the solution becomes permanent pink. Repeat two more times and

complete the table below;-

Table 1

  a) Calculate the average volume of solution K used

  b) The number of moles of solution K reacting

 c) Given that equation for the reaction is:

  MnO_4(aq) + 8H⁺_(aq) + 5Fe²⁺_(aq) Mn²⁺_(aq) + 5 Fe²⁺
_(aq) + 4H₂O_(l)

Determine:

 i) The number of moles of iron (II) salt solution J in 25cm³ of the solution used

 ii) The molarity of solution J

 iii) The concentration of solution J in grams per litre

 d) From your results in C (iii) above, determine:

  i) the value of “n” in the formula (NH₄)₂SO₄FeSO₄nH₂O.

 (N=14, H= 1, S=32, O=16, Fe=56)  

  ii) Correct formula of the iron (II) salt  

iii) The formula mass of the iron (II) salt  

Q2.

1.  120cm³ of solution R

2.  80cm³ of solutions

3.  250cm³ of tap water

4.  25ml or 50ml measuring cylinder

5.  100cm³ glass beaker

6  5 x 5cm piece of white paper

7.  Stop watch or clock.

2.  You are provided with:

 i) Sodium thiosulphate containing 40g/dm³ solution R

 ii) 2M hydrochloric acid solution S

You are to determine the rate of reaction between solution S and the thiosulphate

Procedure:

Measure 20cm³ of solution R into an empty 100cm³ breaker. Place it on a mark ‘X‘ on a white

plain paper. Measure another 20cm³ of solution S. add into R and start off the stop watch. Then

record the time taken for the mark ‘X‘ to become invisible from above. Repeat the procedure by

measuring 17.5cm³ of solution S and adding 2.5cm³ of water and complete the table;-  

Table 2

 a) Draw a graph of reciprocal time (¹/_t) against volume of solution S

 b) Explain the shape of the graph

c) From the graph determine the time taken for the cross ‘X‘ to be invisible at 16.5cm³ of solution S Q3.

1.  Solid Y-1spatulaful

2.  Solid Z-1spatulaful

3.  6 test tubes

4.  1 red + 1blue litmus papers

5.  Metallic spatula

6.  pH paper

3. You are provided with solid Y and Z to carry out the tests below. Write your observations and

inferences in the spaces provided:-

 a) i) Place all solid Y in a clean test tube. Add 10cm³ of distilled water and shake.  

Divide the solution in a (i) above into 4 portions

 ii) To the first portion add sodium hydroxide dropwise until in excess

iii) To the second portion add aqueous ammonia dropwise until in excess

iv ) To the third portion add 5 drops of dilute hydrochloric acid

v) To the fourth portion add 3 drops of lead (II) nitrate solution  

b) i) Scoop a little solid Z on a metallic spatula and heat it over a bunsen flame

  ii) Add all the remaining solid to 10cm³ of distilled water in a test tube and shake.

Divide the solution into 3portions

  iii)to the first portion dip a pH indicator paper

iv) to the second portion add 3 drops of acidified potassium permanganate warm gently  KKC*

v)to the third portion add ½ spatula full of sodium hydrogen carbonate

KAKAMEGA CENTRAL DISTRICT CHEMISTRY PRACTICAL ANSWERS

QUESTION 1 .  

Table 1.

CT  = 1

OP  =1

AC  =1

PA  =1

FA  = 1

5

(a) 22.0 + 22.1 + 21.9 = 22.0cm³

3

Marking points

Complete table (CT) ……….

The table should be completed.

Penalize the following errors if any occurs.

• Arithmetic error in subtraction.
  

• – Values recorded beyond 50cm3
  
• – Inversion of table
  
• Penalize ½ mk only on any one of these errors.
  

Decimal point (d.p) 1mk

All values to be recorded to 1d.p or

All values to be recorded to 2dp second decimal value being 0 or 5 only

Award 0-mark if whole numbers used or 2dp are used.

Accuracy mark (AC)…

Consider any one candidates’ titre if within ± 0.10cm³ of school value award 1mk.

If it is ± 0.11 to 0.20 award ½ mk. If beyond 0.20 award 1mk

Averaging principle (.A)….  

Three titres to be averaged if within ±0.1cm³ to one another.

Two titres can only be arranged if they are consistent.

N/B- If a student averages two titres when three are consistent award 0mk.

Final answer (F. A)…..  

If averaged titre is within 0.0 to 0.10cm³ of S.V award 1mk

0.11 to 0.2cm³ of s.v award ½ mk

If beyond 0.20cm³ award 0mk.

Summary

Complete table (CT) = 1mk

Correct use of decimals(dp) = 1mk

Accuracy (AC) = 1mk

Averaging (PA) = 1mk

Final answer  (FA) = 1mk)

5mks

N/B – school vale (SV) teacher to perform practical to obtain school value.

Calculations

(b) 100cm³ has 0.02moles

   22.0cm³ has- 22x 0.022  1 ½ mk

1000

= 0.00044moles  ½ mk

(c) (i) mole ratio MnO₄ : Fe²⁺ = 1:5

1 mole MnO₄= 5 mol
Fe^{2+  ½ mk}

= 0.00044 x 5

1  

= 0.0022mol  ½ mk

  (ii) 25cm³ has 0.00022mol

1000cm³ has = 1000 X 0.00022

25

= 0.088moldm^-3

(d) (i) RFM of soln has 8.5g

1000cm³ soln = 1000 x 0.85    ½ mk

250

= 34gdm^{-3 ½ mk}

(NH₄)₂ SO₄. FeSO4. nH₂O = 386.4

2(14+1×4) + 32 + 16×4+56 + 32 + 16 x 4 + n(1×2+16) = 386.4

36 + 32 + 64 + 56 + 32 + 64 + 18n = 386.4

284 + 18n = 386.4

28n = 386.4 – 284 ½ mk

n=102.4

18  ½ mk

N=5.6  6  ½ mk  

ii) (NH₄)₂SO₄. FeSO₄. 6H₂O

(iii) R.F.M of J = conc. in gdm^-3

Molarity

= 3.4gdm^{-3  ½ mk}

0.0088mol

= 386.4  ½ mk

Question 2

Table II

Marking points

Complete table (T) ……….  2 ½ mk

Award 1.2 mk for each correct to up to 3 s.f otherwise award 0

CT = 2 ½

DP = ½

AC = ½

Tr = ½

4

Decimal point (dp)………………………  ( ½ mk)

All values of time (t to be whole number or to 1d.p or 2d.p consistently otherwise award 0mk.

Accuracy (AC)…………  ½ mk

Consider time for experiment only if 3 sec of school value (SV) award ½ mk if beyond 0mk.

Trend (Tr)……………  ½ mk

Values of t to be increasing if otherwise 0mk

Summary

Complete table   CT  = 2 ½

Decimal point  DP   = ½

Accuracy Ac  = ½

Trend Tr  = ½

    4mk

1. Graph
   

Labeled axes with correct units = ½ mk

Scale to cover ½ or more of space   = ½

Plotting done correctly = 1

Straight line through 3 point = 1

3mks

1. Straight line graph
   

Increase in concentration; there are more collisions leading to increase in rate of reaction

(c) To read correct value of 1/t from graph  

T=1/t  ½ mk  = ans. ½ mk

Question 3

N/B – Penalize for any contradictory ion  ½ mk

2.  (a)Working out average

Penalties

Wrong arithmetic penalize (- ½ mk)

Correct answer but no working shown ( – ½ mk)

• Value rounded up to 1 d.p ( – ½ mk)

• Accept rounding off of answer to 2d.p
  

(b) moles Na₂CO₃ = 0.05 x 25 = 0.00125 ( ½ mk)

1000

Moles HX = 2x 0.00125 = 0.0025  ( ½ mk)

Molarity of HX = 0.0025 x 1000  ( ½

Titre volume (Av.)

= ……………………

Table 2 and averaging

(c)To be marked as in table 1 bove  5mks

(d) (i) moles B = molarity of HX above x titre volume B

Moles C = moles B

Molarity of C = moles C x 1000

25

(ii) Molarity in d(i) x 56g

(c) Grams KOH in 250ml solution

= ans. In d(ii) ÷ 4……………………………x

Mass KCl in 2.1g = 2.1 – ans. In d(ii) 4

% KCl = 2.1 – x X 100

21

2.  (a) TABLE  

Constant temperature upto 1 ½ min

Then temperature rises slowly to a maximum.

Then remains constant

Lastly it drops slightly

(b) (i) Graph – scale  1mk ( ½ for each axis)

Plot 1mk (for all correct)

For more than ½

Correct ( ½ mk)

Curve 1mk

(ii) Read from graph

(c) Quantity of heat = 40 x 4.2 x temperature change

1000

= ….KJ

(d)  (i) Cu²⁺ + Zn_(s) Zn²⁺_(aq) + Cu_(s)

(ii) Moles Cu²⁺ = 0.2 x 40 = 0.8

1000

= 0.008moles

(iii Ans. in c x 1

0.008

(iv) Some heat is lost into the environment by conduction and convection  

Question 3.    

I

(a)- Jelly solid changes to white solid ( ½ )  

Gas evolved that puts off burning splint   ( ½ )

P is deliquesent ( ½

(b)  (i) White ppt insoluble 1mk

Mg ²⁺ or Ca²⁺ may be present ½

(ii) White ppt formed ½

  Ca²⁺ present

(iii) No white ppt

  Absence of SO^2-₄ or SO^2-₃ ( ½

(iv) White ppt ½

Cl^– present ½

(c) (i) Effervescence occurs/ bubbles (1) and hissing sound

Presence of CO^2-₃ ½

(ii) White ppt insoluble in excess  ½

Mg²⁺ or Ca²⁺ present ½

II

1. Burns with yellow lame ½
   

Inflammable substance or organic

(b) (i) pH is 5-6

Weak acid (H⁺ ions in)

(ii) Effervescence  

– H⁺ ions in Q