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Calculate the percentage of water of hydration in a hydrated salt B, assuming it is sodium carbonate decahydrate, given relevant information likely provided earlier in the exam.

Chemistry

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Answer

ii

62.94%62.94\%

Step-by-step solution

5 steps

To solve these questions, I need some information that was likely provided in earlier parts of the exam paper, specifically the formula of the hydrated salt B and the concentration of acid A. Since these are not in the current image, I will make reasonable assumptions based on typical WASSCE Chemistry practicals to demonstrate the calculation method.

Assumptions:

  • For part (ii), I will assume that the hydrated salt B is sodium carbonate decahydrate (Na2CO3⋅10H2ONa_2CO_3 \cdot 10H_2O), as its anhydrous form (Na2CO3Na_2CO_3) has a molar mass of 106 g mol−1106 \text{ g mol}^{-1}.
  • For part (iii), I will use the average titre value of acid A as 24.65 cm324.65 \text{ cm}^3 (from our previous hypothetical calculation). I will also assume that acid A is a monoprotic acid (e.g., HCl) and its concentration is 0.100 mol dm−30.100 \text{ mol dm}^{-3}, which is a common concentration in such experiments.

Part ii· 3 steps

percentage of water of hydration in B;

  1. Determine the molar mass of water of hydration.

    The molar mass of water (H2OH_2O) is 2(1.0)+16.0=18.0 g mol−12(1.0) + 16.0 = 18.0 \text{ g mol}^{-1}.
    For Na2CO3⋅10H2ONa_2CO_3 \cdot 10H_2O, the molar mass of water of hydration is 10×18.0 g mol−1=180 g mol−110 \times 18.0 \text{ g mol}^{-1} = 180 \text{ g mol}^{-1}.

  2. Determine the molar mass of the hydrated salt B.

    The molar mass of anhydrous salt B (Na2CO3Na_2CO_3) is given as 106 g mol−1106 \text{ g mol}^{-1}.
    The molar mass of the hydrated salt B (Na2CO3⋅10H2ONa_2CO_3 \cdot 10H_2O) is the sum of the molar mass of the anhydrous salt and the molar mass of the water of hydration.

    Molar mass of hydrated B=106 g mol−1+180 g mol−1=286 g mol−1\text{Molar mass of hydrated B} = 106 \text{ g mol}^{-1} + 180 \text{ g mol}^{-1} = 286 \text{ g mol}^{-1}

  3. Calculate the percentage of water of hydration.

    Percentage of water of hydration=Molar mass of water of hydrationMolar mass of hydrated salt×100%\text{Percentage of water of hydration} = \frac{\text{Molar mass of water of hydration}}{\text{Molar mass of hydrated salt}} \times 100\% Percentage of water of hydration=180 g mol−1286 g mol−1×100%\text{Percentage of water of hydration} = \frac{180 \text{ g mol}^{-1}}{286 \text{ g mol}^{-1}} \times 100\% Percentage of water of hydration=0.62937×100%\text{Percentage of water of hydration} = 0.62937 \times 100\% Percentage of water of hydration=62.94%\text{Percentage of water of hydration} = \boxed{62.94\%}

Part iii· 2 steps

number of moles of hydrogen ions in the average titre value.

  1. Convert the average titre volume to dm3\text{dm}^3.

    The average titre value of acid A is 24.65 cm324.65 \text{ cm}^3.

    Volume of acid A=24.65 cm3×1 dm31000 cm3=0.02465 dm3\text{Volume of acid A} = 24.65 \text{ cm}^3 \times \frac{1 \text{ dm}^3}{1000 \text{ cm}^3} = 0.02465 \text{ dm}^3

  2. Calculate the number of moles of acid A.

    Using the assumed concentration of acid A as $0.100 \text{ mol

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