Provide calculation steps and general answers for titration and experimental data analysis, assuming reasonable values for missing data.

I will provide the steps and general answers for the questions. Since the tables are blank, I will outline the calculation methods and make reasonable assumptions for observations in Question 2 to demonstrate the process. You will need to substitute your actual experimental data into the formulas for numerical answers.

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QUESTION ONE

Calculations

• Assumption for Table 1 (Mass of solid A):

  • Mass of weighing bottle + solid A = $15.25 \text{ g}$
  • Mass of weighing bottle alone = $9.10 \text{ g}$
  • Mass of solid A = $15.25 \text{ g} - 9.10 \text{ g} = 6.15 \text{ g}$ (This value is within the specified range of 6.0 to 6.25 g)

• Assumption for Table 3 (Titration readings):

  • Approximate titre = $25.00 \text{ cm}^3$
  • First accurate titre = $24.60 \text{ cm}^3$
  • Second accurate titre = $24.50 \text{ cm}^3$
  • Mean titre = $\frac{24.60 + 24.50}{2} = 24.55 \text{ cm}^3$

1) Using the titre, calculate the molarity of solution A and then solution B.

• Step 1: Calculate the molar mass of $\text{Na}_2\text{S}_2\text{O}_3 \cdot 5\text{H}_2\text{O}$ (solid A).

  $$M_r(\text{Na}_2\text{S}_2\text{O}_3 \cdot 5\text{H}_2\text{O}) = (2 \times 22.99) + (2 \times 32.07) + (3 \times 16.00) + (5 \times (2 \times 1.008 + 16.00))$$ $$M_r(\text{Na}_2\text{S}_2\text{O}_3 \cdot 5\text{H}_2\text{O}) = 45.98 + 64.14 + 48.00 + (5 \times 18.016)$$ $$M_r(\text{Na}_2\text{S}_2\text{O}_3 \cdot 5\text{H}_2\text{O}) = 45.98 + 64.14 + 48.00 + 90.08 = 248.20 \text{ g/mol}$$

• Step 2: Calculate the moles of $\text{Na}_2\text{S}_2\text{O}_3 \cdot 5\text{H}_2\text{O}$ in $250 \text{ cm}^3$ of solution A.

  $$\text{Moles of A} = \frac{\text{Mass of A}}{M_r(\text{Na}_2\text{S}_2\text{O}_3 \cdot 5\text{H}_2\text{O})}$$ $$\text{Moles of A} = \frac{6.15 \text{ g}}{248.20 \text{ g/mol}} = 0.02478 \text{ mol}$$

• Step 3: Calculate the molarity of solution A.

  $$\text{Molarity of A} = \frac{\text{Moles of A}}{\text{Volume of solution A (in L)}}$$ $$\text{Molarity of A} = \frac{0.02478 \text{ mol}}{0.250 \text{ L}} = 0.09912 \text{ mol/L}$$

• Step 4: Calculate the moles of $\text{S}_2\text{O}_3^{2-}$ used in the titration.

  $$\text{Moles of S}_2\text{O}_3^{2-} = \text{Molarity of A} \times \text{Mean titre (in L)}$$ $$\text{Moles of S}_2\text{O}_3^{2-} = 0.09912 \text{ mol/L} \times \frac{24.55}{1000} \text{ L} = 0.002433 \text{ mol}$$

• Step 5: Calculate the moles of $\text{I}_2$ reacted. From the equation: $2\text{S}_2\text{O}_3^{2-} + \text{I}_2 \longrightarrow \text{S}_4\text{O}_6^{2-} + 2\text{I}^-$ The mole ratio $\text{S}_2\text{O}_3^{2-} : \text{I}_2$ is $2:1$.

  $$\text{Moles of I}_2 = \frac{\text{Moles of S}_2\text{O}_3^{2-}}{2}$$ $$\text{Moles of I}_2 = \frac{0.002433 \text{ mol}}{2} = 0.0012165 \text{ mol}$$

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