Question

A 500 mL of Mercury solution was prepared by mixing 2.45g of Mercury chloride and 4.68g of Mercury sulfate in one container them dissolve and complete to volume with distilled water. What is the final concentration Mercury in the final solution? Express concentration in Molarity to 4 s.f and %(w/v) to 3 s.f.

Answer 1

To calculate the final concentration of mercury in the solution, we need to first determine the total mass of mercury contributed by both mercury chloride (HgCl₂) and mercury sulfate (HgSO₄).

Step 1: Calculate the mass of mercury from each compound

1. Mercury chloride (HgCl₂)

• Molar mass of HgCl₂:

  • Hg: 200.59 g/mol
  • Cl: 35.45 g/mol × 2 = 70.90 g/mol
  • Total: 200.59 + 70.90 = 271.49 g/mol

• Moles of HgCl₂:

  • Moles = mass / molar mass
  • Moles of HgCl₂ = 2.45 g / 271.49 g/mol ≈ 0.00903 mol

• Each mole of HgCl₂ contains 1 mole of mercury (Hg).

• Therefore, moles of mercury from HgCl₂ = 0.00903 mol.

2. Mercury sulfate (HgSO₄)

• Molar mass of HgSO₄:

  • Hg: 200.59 g/mol
  • S: 32.07 g/mol
  • O: 16.00 g/mol × 4 = 64.00 g/mol
  • Total: 200.59 + 32.07 + 64.00 = 296.66 g/mol

• Moles of HgSO₄:

  • Moles = mass / molar mass
  • Moles of HgSO₄ = 4.68 g / 296.66 g/mol ≈ 0.01577 mol

• Each mole of HgSO₄ contains 1 mole of mercury (Hg).

• Therefore, moles of mercury from HgSO₄ = 0.01577 mol.

Step 2: Total moles of mercury in the solution

Total moles of mercury = Moles from HgCl₂ + Moles from HgSO₄

• Total moles of mercury = 0.00903 mol + 0.01577 mol = 0.02480 mol.

Step 3: Calculate the molarity of mercury in the final solution

The final volume of the solution is 500 mL, which is 0.500 L.

Molarity (M) = Moles of solute / Volume of solution in liters

• Molarity of mercury = 0.02480 mol / 0.500 L = 0.0496 M.

Step 4: Calculate the percentage concentration (w/v)

To find the % (w/v), we first calculate the total mass of mercury from the compounds:

Path 1: Find the total mass of mercury directly.

• From HgCl₂: 0.00903 mol × 200.59 g/mol = 1.81 g of Hg.
• From HgSO₄: 0.01577 mol × 200.59 g/mol = 3.17 g of Hg.
• Total mass of mercury = 1.81 g + 3.17 g = 5.00 g.

Path 2: % (w/v) = (mass of solute in grams / volume of solution in mL) × 100

• % (w/v) = (5.00 g / 500 mL) × 100 = 1.00 % (w/v).

Final Results

• Molarity of mercury in the final solution = 0.0496 M (to 4 significant figures).
• Concentration of mercury in % (w/v) = 1.00 % (to 3 significant figures).