To find the final concentration of mercury in the solution, we will first need to determine the amount of mercury contributed by both mercury chloride (HgCl₂) and mercury sulfate (HgSO₄) in terms of moles, then convert into molarity and percent concentration.
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Calculate the Molar Masses:
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Mercury Chloride (HgCl₂):
- Molar mass of Hg = 200.59 g/mol
- Molar mass of Cl = 35.45 g/mol (2 Cl)
- Molar mass of HgCl₂ = 200.59 g/mol + 2 × 35.45 g/mol = 200.59 + 70.90 = 271.49 g/mol
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Mercury Sulfate (HgSO₄):
- Molar mass of S = 32.07 g/mol
- Molar mass of O = 16.00 g/mol (4 O)
- Molar mass of HgSO₄ = 200.59 g/mol + 32.07 g/mol + 4 × 16.00 g/mol = 200.59 + 32.07 + 64.00 = 296.66 g/mol
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Calculate the moles of each compound:
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Moles of HgCl₂: \[ \text{Mass} = 2.45 , \text{g} \] \[ \text{Moles} = \frac{\text{Mass}}{\text{Molar mass}} = \frac{2.45 , \text{g}}{271.49 , \text{g/mol}} \approx 0.00903 , \text{mol} \] Each mole of HgCl₂ provides 1 mole of Hg, so moles of Hg from HgCl₂ = 0.00903 mol.
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Moles of HgSO₄: \[ \text{Mass} = 4.68 , \text{g} \] \[ \text{Moles} = \frac{\text{Mass}}{\text{Molar mass}} = \frac{4.68 , \text{g}}{296.66 , \text{g/mol}} \approx 0.01575 , \text{mol} \] Each mole of HgSO₄ provides 1 mole of Hg, so moles of Hg from HgSO₄ = 0.01575 mol.
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Total moles of mercury: \[ \text{Total moles of Hg} = \text{moles from HgCl₂} + \text{moles from HgSO₄} = 0.00903 , \text{mol} + 0.01575 , \text{mol} \approx 0.02478 , \text{mol} \]
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Calculate the final concentration in Molarity: The final volume of the solution is 50 mL, which is 0.050 L. \[ \text{Molarity (M)} = \frac{\text{Total moles of Hg}}{\text{Volume in L}} = \frac{0.02478 , \text{mol}}{0.050 , \text{L}} = 0.4956 , \text{M} \]
Rounded to 4 significant figures: \[ \text{Molarity} = 0.4956 , \text{M} \]
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Calculate % (w/v):
- % (w/v) is calculated as: \[ \text{Percent (w/v)} = \left( \frac{\text{Mass of solute (g)}}{\text{Volume of solution (mL)}} \right) \times 100% \] The total mass of mercury from both sources:
- From HgCl₂:
- Mass of Hg from HgCl₂ = \( 2.45 , \text{g} \times \frac{200.59}{271.49} \approx 1.81 , \text{g} \)
- From HgSO₄:
- Mass of Hg from HgSO₄ = \( 4.68 , \text{g} \times \frac{200.59}{296.66} \approx 3.16 , \text{g} \)
The total mass of Mercury is: \[ \text{Total mass of Hg} = 1.81 + 3.16 = 4.97 , \text{g} \]
Now calculate the % (w/v): \[ \text{Percent (w/v)} = \left( \frac{4.97 , \text{g}}{50 , \text{mL}} \right) \times 100 = 9.94 % \]
Rounded to 3 significant figures: \[ \text{Percent (w/v)} = 9.94% \]
Final Results:
- Concentration in Molarity: 0.4956 M
- Concentration in % (w/v): 9.94%