Chemistry Alternative To Practical 2023 Cover

Answer all the questions

1. A is a solution containing 4.00 g of the acid, HX, per dm³.
   
   B contains 1.00 g sodium hydroxide per 250 cm³ of solution.
   
   A was titrated with 20.0 cm³ portions of B using methyl orange as indicator and the following results were obtained:

   Burette Readings / cm3	1	2	3
   Final reading	23.10	24.60	46.30
   Initial reading		1.50
   Volume of A used	23.10		23.20

   
   

   (a)

   1. Copy and complete the table.
   
   
   2. Calculate the average volume of A used.
   
   
   3. State the colour of the indicator:
      - in solution B before the titration;
      - at the end-point.

   
   

   (b) From the information provided, calculate the:

   1. Concentration of B in mol·dm⁻³
   
   
   2. Concentration of A in mol·dm⁻³
   
   
   3. Molar mass of the acid HX

   
   

   The reaction equation is:
   
   HX_(aq) + NaOH_(aq) → NaX_(aq) + H₂O_(l)

   Relative Atomic Masses:
   
   H = 1.0, O = 16.0, Na = 23.0

   [18 marks]

   
   
   questions 1 (a) Answer

   (a)

   1. Completed Table:
      
      

      Burette Readings / cm3	1	2	3
      Final reading	23.10	24.60	46.30
      Initial reading	0.00	1.50	23.10
      Volume of A used	23.10	23.10	23.20

   
   
   2. Average volume of A used:
      
      Using the consistent values from Titration 1, 2, and 3:
      
      $$\text{Average}=\frac{23.10+23.10+23.20}{3}$$
      $$=\frac{69.40}{3}$$
      

           = 23.13 cm³

   
   
   3. Colour of the indicator:
      
      - In solution B before the titration: Yellow
      
      - At the end-point: Orange
      
      
      

   questions 1 (b) Answer

   (b)

   1. Concentration of B in mol·dm⁻³:
      
      Given:
      Mass of NaOH = 1.00 g
      Volume of solution = 250 cm³ = 0.250 dm³
      Molar mass of NaOH = 23 (Na) + 16 (O) + 1 (H) = 40 g/mol
      
      Number of moles of NaOH =
      $$\text{mol}=\frac{1.00}{40}$$ = 0.025 mol
      
      Concentration = $$\text{mol·dm}⁻3=\frac{0.025}{0.250}$$ = 0.10 mol·dm⁻³
   
   
   2. Concentration of A in mol·dm⁻³:
      
      From the reaction: HX + NaOH → NaX + H₂O
      This is a 1:1 molar ratio.
      
      Volume of B used = 20.0 cm³ = 0.0200 dm³
      Concentration of B = 0.10 mol·dm⁻³
      Moles of NaOH used = 0.10 × 0.0200 = 0.002 mol
      
      Therefore, moles of HX in 23.13 cm³ of A = 0.002 mol
      Volume of A used = 23.13 cm³ = 0.02313 dm³
      
      Concentration of A = $$\text{mol·dm}⁻3=\frac{0.002}{0.02313}$$ ≈ 0.0865 mol·dm⁻³
   
   
   3. Molar mass of HX:
      
      Mass of HX in 1 dm³ of solution = 4.00 g
      Moles of HX in 1 dm³ = 0.0865 mol (from part ii)
      
      Molar mass = $$\text{g·mol}⁻1=\frac{4.00}{0.0865}$$ ≈ 46.24 g/mol
      
      Since H = 1.0, the molar mass of X ≈ 45.24 g/mol → likely element is Cl (35.5), so the acid is:
      HX = HCl



2. (a) C is a mixture of two inorganic salts. The following tests were carried out on C. Copy and complete the table.

   
   

   	Test	Observation	Inference
   (a) (i)	C + distilled water + filtration	Partly dissolves to give a colourless solution and a white residue
   (ii)	Filtrate + NaOH(aq) in drops then in excess	White gelatinous precipitate formed Soluble in excess
   (iii)	Filtrate + NH3(aq) in drops then in excess	Precipitate dissolves
   (iv)	Filtrate + dil HNO3 + AgNO3(aq) + NH3(aq)		Cl− confirmed
   (v)	Residue + dil HCl + heat	CO2 gas evolved

   [12 marks]

   
   

   (b) Copy and complete the following table stating what would be observed in each case if the test were performed as indicated.

   
   

   	Test	Observation
   (b)(i)	Pb(NO3)2 + distilled water
   (ii)	Portion from (b)(i) + NaOH(aq) in drops then in excess
   (iii)	Portion from (b)(i) + NH3(aq) in drops then in excess
   (iv)	Portion from (b)(i) + KI

   [6 marks]

   
   
   questions 2 (a) Answer

   (a) completed Table.

   
   

   	Test	Observation	Inference
   (a) (i)	C + distilled water + filtration	Partly dissolves to give a colourless solution and a white residue	C contains both soluble and insoluble salts
   (ii)	Filtrate + NaOH(aq) in drops then in excess	White gelatinous precipitate formed Soluble in excess	Al3+ present
   (iii)	Filtrate + NH3(aq) in drops then in excess	Precipitate dissolves	Al3+ confirmed
   (iv)	Filtrate + dil HNO3 + AgNO3(aq) + NH3(aq)	White precipitate formed, dissolves in excess NH3	Cl− confirmed
   (v)	Residue + dil HCl + heat	CO2 gas evolved	CO32− present; residue is likely a carbonate

   
   
   
   

   questions 2 (b) Answer
   
   (b) Complete table with stated Observation under given test performed.

   
   

   	Test	Observation
   (b)(i)	Pb(NO3)2 + distilled water	Colourless solution formed
   (ii)	Portion from (b)(i) + NaOH(aq) in drops then in excess	White precipitate formed Insoluble in excess
   (iii)	Portion from (b)(i) + NH3(aq) in drops then in excess	White precipitate formed Insoluble in excess
   (iv)	Portion from (b)(i) + KI	Bright yellow precipitate formed





3. (a) Name a suitable drying agent for each of the following gases:

   1. NH₃;
   
   
   2. CO₂.

   [2 marks]

   questions 3 (a) Answer
   
   

   Gas	Drying Agent
   NH3	Quicklime (CaO)
   CO2	Concentrated H2SO4

   
   
   

   (b) State two chemical tests that could be used to show that a sample of bread contains starch and protein.

   [2 marks]

   questions 3 (b) Answer
   

   Test for Starch:
   
   Add iodine solution to the bread sample.
   
   Result: A blue-black coloration confirms the presence of starch.

   Test for Protein:
   
   Add Biuret reagent to the sample.
   
   Result: A purple or violet coloration indicates the presence of protein.

   
   
   

   (c) Explain briefly each of the following statements:
   
   

   • (i) HCl, though a covalent compound, conducts electricity in solution;
   
   
   
   • (ii) Dry sodium chloride crystals with traces of calcium chloride get hydrated when exposed to the atmosphere.

   [4 marks]

   
   
   Question 3 (c) Answer
   

   (i) HCl conducts electricity in solution:
   
   Although HCl is a covalent compound, when dissolved in water, it ionizes completely to produce H⁺ and Cl⁻ ions.
   
   Result: The free ions in solution allow the solution to conduct electricity.

   (ii) Hydration of NaCl crystals with CaCl₂ traces:
   
   Calcium chloride is a hygroscopic substance, meaning it readily absorbs moisture from the air.
   
   Result: This causes the dry sodium chloride crystals to become hydrated when exposed to the atmosphere.

   
   
   
   

   (d) Name a liquid substance that could be used in the laboratory for:
   
   (i) dissolving dry mortar on floor tiles;
   
   (ii) removing KMnO₄ stains;
   
   (iii) drying acid anhydrides.

   [3 marks]

   
   
   Question 3 (d) Answer
   
   
   

   	Purpose	Liquid Substance
   i	Dissolving dry mortar on floor tiles	Hydrochloric acid (HCl)
   ii	Removing KMnO4 stains	Oxalic acid solution
   iii	Drying acid anhydrides	Concentrated sulfuric acid (H2SO4)

   
   
   

   (e) A solution of HCl was titrated with sodium hydroxide solution.
   
   State three precautions that are necessary when carrying out the experiment.

   [3 marks]

   
   Question 3 (e) Answer
   
   

   Precautions:

   • Rinse all apparatus (burette, pipette, conical flask) with their respective solutions before use.
   
   
   • Ensure the burette is clamped vertically and free of air bubbles before taking readings.
   
   
   • Add the acid gradually near the end point while swirling the conical flask to avoid overshooting the end point.
   
   
   
   
   

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