Chemistry Paper 3 Questions and Answers - Asumbi Girls Mock Examinations 2022

QUESTIONS

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 (3 marks)

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 25c m³ 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)

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) + X₂CO₃(s) → 2XCl(aq) + CO₂(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)

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)

3. You’re provided with solid D. Carry out the tests shown below on the solid.
   1. Heat a spatula full of D in A clean dry test – tube.
      

      OBSERVATIONS	INFERENCES

   2. Put a spatula end- full of D in a boiling tube. Half fill it with water. shake this mixture
      

      OBSERVATIONS	INFERENCES

   3. Divide the resultant mixture in (b) above into 5 portions
      

      OBSERVATIONS	INFERENCES

      1. To the first portion add dilute nitric acid followed by a few drops of Barium nitrate
         

         OBSERVATIONS	INFERENCES

      2. To the second portion, add nitric acid a few drops followed by lead (II) nitrate and then warm the mixture.
         

         OBSERVATIONS	INFERENCES

      3. To the third portion, add sodium hydroxide solution drop wise until in excess. Warm this mixture. Test any gas produced withy Litmus paper
         

         OBSERVATIONS	INFERENCES

   4. You are provided with liquid B. Carry out the tests shown below and write your observations and inferences in the spaces provided.
      1. To about 1cm³ of liquid B in a test – tube, add about 1cm3 of distilled water and shake the mixture.
         

         OBSERVATIONS	INFERENCES

      2. To about 1cm³ of liquid B in a test tube add a small amount of solid sodium hydrogen carbonate
         

         OBSERVATIONS	INFERENCES

      3. To about 2cm³ of liquid B in A test – tube, add about 1cm3 of acidified potassium dichromate (VI). Warm the mixture gently and allow it to stand for about one minute.
         

         OBSERVATIONS	INFERENCES

CONFIDENTIAL
Each student requires

• 80 cm³ of solution A prepared by dissolving 4.0g of NaOH in water made up to 1 litre.
• 200cm³ of solution B which is 0.1M HCl.
• 2.5 of a salt mixture prepared by mixing 1.5 g of sodium carbonate (Anhydrous) and 1.0g of sodium chloride.
• One burette (50ml)
• One 25cm³ pipette
• Pipette filler
• Complete stand
• Filter funnel
• White tile
• 3 conical flasks
• 250ml volumetric flask.
• Labels (6)
• 500ml distilled water in a wash bottle.
• 200cm³ of solution S₃ which is sodium thiosulphate of concentration 0.2 M
• 80cm of 2m hydrochloric acid solution
• 100ml empty glass beaker.
• One stop watch/ clock.
• white piece paper
• One 50ml measuring cylinder
• About 2g of solid D (A mixture of ammonium sulphate and zinc sulphate in the ratio 1:1).
• About 5cm³ of liquid B which is absolute ethanol
• Two red litmus paper.
• One metallic spatula.
• About 1g of solid sodium hydrogen carbonate
• Test- tube holder.
• One boiling tube
• Seven clean dry test tubes.

Provide access to:

• Means of heating
• Phenolphthalein indicator
• Methyl - indicator
• Acidified potassium dichromate (vi) solution.
• 2m nitric (V) acid solution
• 2m Barium nitrate solution
• 2m ammonia solution.
• 2m sodium hydroxide solution.
• 2m lead nitrate solution

Note: The solution should be supplied with droppers.

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
      = ⁴/₄₀ = 0.1molar
      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
2. 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. 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	20
   B	40	10	10	25
   C	30	20	10	35
   D	20	30	10	53
   E	10	40	10	103

   
   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