To find the final concentration of mercury in the solution, we first need to determine the total amount of mercury from both mercury chloride (HgCl₂) and mercury sulfate (HgSO₄).
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Calculating the molar mass of mercury compounds:
- Mercury chloride (HgCl₂):
- Hg: 200.59 g/mol
- Cl: 35.45 g/mol (2 Cl)
- Molar mass of HgCl₂ = 200.59 + 2(35.45) = 200.59 + 70.90 = 271.49 g/mol
- Mercury sulfate (HgSO₄):
- Hg: 200.59 g/mol
- S: 32.07 g/mol
- O: 16.00 g/mol (4 O)
- Molar mass of HgSO₄ = 200.59 + 32.07 + 4(16.00) = 200.59 + 32.07 + 64.00 = 296.66 g/mol
- Mercury chloride (HgCl₂):
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Calculate the moles of mercury in each compound:
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From mercury chloride: \[ \text{Moles of HgCl₂} = \frac{2.45, \text{g}}{271.49, \text{g/mol}} \approx 0.00903, \text{mol} \] Since each molecule of HgCl₂ contains 1 mercury atom: \[ \text{Moles of Hg from HgCl₂} = 0.00903, \text{mol} \]
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From mercury sulfate: \[ \text{Moles of HgSO₄} = \frac{4.68, \text{g}}{296.66, \text{g/mol}} \approx 0.01576, \text{mol} \] Since each molecule of HgSO₄ also contains 1 mercury atom: \[ \text{Moles of Hg from HgSO₄} = 0.01576, \text{mol} \]
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Total moles of mercury in the solution: \[ \text{Total moles of Hg} = 0.00903 + 0.01576 = 0.02479, \text{mol} \]
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Calculating the molarity of the final solution: The total volume of the solution is 500 mL, which is 0.500 L. \[ \text{Molarity (M)} = \frac{\text{Moles of solute}}{\text{Volume of solution in liters}} = \frac{0.02479, \text{mol}}{0.500, \text{L}} = 0.04958, \text{M} \] Expressed to four significant figures: \[ \text{Final Molarity} = 0.04958, \text{M} \]
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Calculating % (w/v) concentration: To find the % (w/v), we need to express the mass of mercury in 100 mL.
- The total mass of mercury from both salts:
- From HgCl₂: 2.45 g contains 2.45 g of HgCl₂ and yields 2.45 g (calculated previously).
- From HgSO₄: 4.68 g contains 4.68 g of HgSO₄ and yields 4.68 g (calculated previously).
- The total mass of mercury: \[ \text{Total mass of mercury from HgCl}_2 = 2.45, \text{g} \times \frac{200.59}{271.49} \approx 1.80, \text{g} \] \[ \text{Total mass of mercury from HgSO}_4 = 4.68, \text{g} \times \frac{200.59}{296.66} \approx 3.16, \text{g} \] Adding these together gives: \[ \text{Total mass of mercury} \approx 1.80, \text{g} + 3.16, \text{g} = 4.96 g \]
Since the volume of the final solution is 500 mL, the % (w/v) concentration is calculated as: \[ \text{% (w/v)} = \frac{\text{mass of solute (g)}}{\text{volume of solution (mL)}} \times 100 = \frac{4.96, \text{g}}{500, \text{mL}} \times 100 = 0.992, % \] Expressed to three significant figures: \[ \text{Final % (w/v)} = 0.992 , % \]
- The total mass of mercury from both salts:
Final Results:
- Concentration (Molarity): \( 0.04958 , \text{M} \)
- Concentration (% w/v): \( 0.992 , % \)