Chemistry Paper 3 Questions and Answers - Baringo North Joint Evaluation Mock Exams 2022

Questions

Instructions to candidates.

• Answer ALL question in the spaces provided in the question paper.
• You are not allowed to start working with the apparatus for the first 15 minutes of the 2¼ hours allowed for this paper. This time is to enable you read the question paper and make sure you have all the chemicals and apparatus you may need.
• Mathematical tables and silent electronic calculators may be used.
• All workings MUST be clearly shown where necessary.
  
  

QUESTION 1
You are provided with:

• Solution A, containing 4.0gdm-3 of sodium hydroxide
• Solution B, hydrochloric acid
• 2.5 g of a mixture of two salts, XCl (RFM 58.5) and X2CO3 (RFM 106)
  You are required to:
  
  1. Standardize solution B, hydrochloric acid.
  2. Determine the mass composition of the salt mixture

PROCEDURE I

1. Fill the burette with solution B
2. Pipette 25 cm³ of solution A into a clean dry conical flask. Then add 2 -3 drops of phenolphthalein indicator.
3. Titrate solution A solution with solution B. Record your results in the table below.
4. 
   Repeat the procedure two more times to retain concord and values.
   TABLE 1
   

   Titration number	1	2	3
   Final burette reading (cm3)
   Initial burette reading(cm3)
   Volume of acid used (cm3)

   
   
   1. Calculate the average volume of solution B used. (1mark)
   2. Find;
      1. Moles of sodium hydroxide that reacted with the acid. (2 marks)
      2. Moles of hydrochloric acid present in the average volume. (1mark)
      3. Molarity of the acid (1mark)

PROCEDURE II

1. Put about 100cm³ of water in a 250ml volumetric flask add all the 2.5g of salt mixture. Shake the mixture to dissolve and the solid. Top up the solution to the mark with distilled water Label this solution C
2. Fill this burette with solution B.
3. Pipette 25cm³ of solution C and put it into a clean conical flask. Add 3 drops of methyl orange indicator.
4. Titrate solution C with solution B. Record your results in the table below.
5. Repeat the titration two more times
   TABLE II (3 marks)
   

   TITRATION	1	2	3
   Final burette reading (cm3)
   Initial burette reading (cm3)
   Volume of solution B used (cm3)

   3. Calculate the average volume of solution B (1mark)
   4. Calculate the number of moles in the hydrochloric acid used (1mark)
   5. The equation for the reaction of the acid with one of the salts in the mixture is:
      2HCl_(aq) + X₂CO_3(s) → 2XCl_(aq) + CO_2(g) + H₂O_(l)
      Calculate;
      1. Moles of X₂CO₃ that reacted with the acid in the experiment (1mark)
      2. Molarity of X₂CO₃ (2 marks)
   6. Calculate the mass of the salt mixture in gdm^-3. (1mark)
   7. Calculate the percentage of XCL in this mixture (2 marks)

QUESTION 2

In this experiment, you’re required to determine the time takes for a precipitate to be formed when S3 which is sodium thiosulphate solution, reacts with dilute hydrochloric acid.

PROCEDURE

1. Using a measuring cylinder measure 50cm³ of S3 into a 100ml beaker.
2. Make a pencil cross on a white piece of paper so that when a beaker is placed top of the paper , the cross can be seen through the bottom of the beaker.
3. To solution A add 10 cm³ of 2M hydrochloric acid and at the same time start a stop watch / stop clock. Swirl the contents of the beaker twice and then place it over the cross on the paper. Look at the cross from above the beaker through the mixture. Stop the stop watch immediately the precipitate makes the cross invisible. Record time taken for the cross to become invisible in the table below, rinse beaker.
4. Repeat the procedure with solutions B, C, D and E as per the table.
   

   SOLUTION	Volume of solution S3 in the beaker (cm3)	Volume of water added (cm3)	Volume of 2M HCl	Time taken in seconds
   A	50	0	10
   B	40	10	10
   C	30	20	10
   D	20	30	10
   E	10	40	10

   
   
   1. Plot the graph of volume of solution S3 (y – axis) against time (4 marks)
   2. From the graph state the relationship between concentration of solution S3 and time. (1mark)
   3. Why is water added to the solution S3? (1mark)

QUESTION 3

You’re provided with solid D. Carry out the tests shown below on the solid.

 
(d) You are provided with liquid B. Carry out the tests shown below and write your observations and inferences in the spaces provided.

Marking Scheme 

Table 1

Marking

• Complete table award; ✔
• Decimal consistency; ✔
• Accuracy 0.1;✔
• School value; ✔
  Principles of averaging:
  Average volume = 24.5 + 24.5 + 24.5 = 24.5; ✔ (½ mark)
                                               3
  = 24.5 cm³; ✔(½ mark)

1.        
   1. Moles of sodium hydroxide used
      Molarity of solution:
      Moles =     Mass /litre   
                        RMM
      =       4      = 0.1molar
             40
      If 1000 cm³ → 0.1mole
      Then 25 cm³ → 25 x 0.1 = 0.0025 moles;
                                    1000
      
      
   2. Moles of hydrochloric acid
      NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)
      Mole ratio = 1:1;
      Thus moles of acid = 0.0025 moles;
      
      
   3. Molarity of acid.
      Volume of acid reacting = average titre in (a) e.g. 24.5 cm3
      If 24.5 cm3 → 0.0025 moles
      Then 1000 cm3 → 1000 x 0.0025 = 0.1020 molar;
                                            24.5

Table II
Marking:
Complete table ✔ - 1mark
Decimal consistency ✔- 1 mark
Accuracy ✔ - 1mark
School value ✔- 1mark
Expected titre = 28.3 cm3

3. Average volume
   28.3 + 28.3 + 28.3 = 28.3 cm3
              3
   All 3 values within 0.1 of each other and used – 1mark
   Only 2 within 0.1 of each other and used – ½ mark
   Inconsistent value used – 0 mark
   
   
4. Answer in (b) (iii) x average volume
                    1000
   Example:
   Molarity of the acid calculated in (b) (iii) = 0.102 molar.
   Thus 1000 cm3 → 0.102 moles
   28.3 cm3 → 28.3 x 0.102 = 0.0028866 moles;
                               1000
5.          
   1. Moles of carbonate that reacted:
      Using mole ratio, 2 moles of acid reacts with 1 mole of carbonate
      Thus moles of carbonate reacting = answer in (d) x 1
                                                                       2
      Example:
      2 moles of acid → 1 mole of carbonate
      Thus 0.0028866 moles of acid → 0.0028866 x 1
                                                                   2
      = 0.0014433 moles;
      
      
   2. Molarity of carbonate
      25 cm3 of carbonate → 0.0014433
      Then 1000 cm3 → 1000 x 0.001443 = 0.057732 molar
                                                25
6. Mass of the salt mixture in gdm-3
   250 cm³ → 2.5g
   1000 cm³ → 1000 x 2.5 = 10g;
                              250
7. Percentage of XCl in the mixture
   Mass of X₂CO₃ in 1 litre
   Molarity = 0.057732 molar / answer in (e) (ii).
   Mass = 0.057732 x 106 = 6.119592g;
   Mass of XCl = 10 – 6.119592
   = 3.880408;
   Percentage = 3.880408 x 100
                               10
   = 30.80408%
   Note: Use school based values.

Question 2

1. Marking:
   • Complete table✔
   • Decimal consistency ✔
   • Trend (increasing time) ✔
   • School value ✔
     
     Graph Trend:
     
     
     
     Marking:
     Scale – ½ mark
     Axes – ½ mark
     Plotting – 1mark
     Curve – 1mark
2. As the concentration decreases ,the time increases ✔
3. To keep the column of solution constant through the experiment ✔

Question 3 (a)