To calculate the concentration of mercury(II) iodide (HgI2) in moles per liter (mol/L), we can follow these steps:
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Determine the molar mass of mercury(II) iodide (HgI2):
- The molar mass of mercury (Hg) is approximately 200.59 g/mol.
- The molar mass of iodine (I) is approximately 126.90 g/mol.
Molar mass of HgI2 = 200.59 g/mol (for Hg) + 2 × 126.90 g/mol (for I) \[ = 200.59 , \text{g/mol} + 253.80 , \text{g/mol} = 454.39 , \text{g/mol} \]
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Convert the mass of HgI2 to moles: We have 0.0118 g of HgI2. \[ \text{Moles of HgI2} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} = \frac{0.0118 , \text{g}}{454.39 , \text{g/mol}} \] \[ \approx 2.59 \times 10^{-5} , \text{mol} \]
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Calculate the concentration (in mol/L): The volume of the solution is 250 mL, which is 0.250 L. \[ \text{Concentration (C)} = \frac{\text{moles of solute}}{\text{volume of solution in L}} = \frac{2.59 \times 10^{-5} , \text{mol}}{0.250 , \text{L}} \] \[ = 1.036 \times 10^{-4} , \text{mol/L} \]
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Expressing the final answer with the correct significant figures: The value 0.0118 g has three significant figures. Therefore, we should round our concentration to three significant figures as well. \[ \text{Concentration} = 1.04 \times 10^{-4} , \text{mol/L} \]
Thus, the concentration of the mercury(II) iodide solution is \( \boxed{1.04 \times 10^{-4}} , \text{mol/L} \).
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