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The procedure of a first-hand investigation conducted in a school laboratory to determine the percentage of sulfate in a lawn fertiliser is shown - HSC - SSCE Chemistry - Question 29 - 2015 - Paper 1

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Question 29

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The procedure of a first-hand investigation conducted in a school laboratory to determine the percentage of sulfate in a lawn fertiliser is shown. 2.00 g of a sampl... show full transcript

Worked Solution & Example Answer:The procedure of a first-hand investigation conducted in a school laboratory to determine the percentage of sulfate in a lawn fertiliser is shown - HSC - SSCE Chemistry - Question 29 - 2015 - Paper 1

Step 1

Suggest modifications that could be made to the procedure to improve the results of this investigation. Justify your suggestions.

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Answer

  1. Heating the Mixture: After adding barium chloride, the mixture could be heated for a longer duration. This would allow fine BaSO₄ particles to agglomerate into larger clusters, making them more likely to be trapped by the filter paper.

  2. Use of Sintered Glass Filter: To improve the filtration process, a sintered glass filter could be used. This type of filter has smaller pores that can effectively trap finer precipitate particles, leading to better recovery of BaSO₄.

  3. Multiple Samples: Instead of using a single sample of the fertiliser, multiple samples should be collected and treated. This practice will result in a more thorough understanding of the sulfate content, as variations in composition can exist in different batches of fertiliser.

Step 2

Calculate the percentage of sulfate in the original fertiliser sample.

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Answer

  1. Calculations of Molar Mass:

    • Barium sulfate (BaSO₄): Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s)
    • Molar mass of BaSO₄ = 137.33 (Ba) + 32.07 (S) + 4 × 16.00 (O) = 233.39 g/mol
  2. Mass of Sulfate in Precipitate:

    • Mass of precipitate collected = 2.23 g
    • Mass of SO₄²⁻ in BaSO₄ = (96.07/233.39) × 2.23 = 0.91 g
  3. Percentage Calculation:

    • Percentage of sulfate in fertiliser:

    rac{0.91 ext{ g}}{2.00 ext{ g}} imes 100 = 45.95 ext{%}

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