Question

A chemist prepares a solution of mercury(II) iodide (HgI2) by measuring out 0.0118 g of mercury(II) iodide into a 250. mL volumetric flask and filling the flask to the mark with water.
Calculate the concentration in mol/L of the chemist's mercury(II) iodide solution. Be sure your answer has the correct number of significant digits.

Answer 1

To calculate the concentration of mercury(II) iodide (HgI2) in moles per liter (mol/L), we can follow these steps:

1. **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)
\[
= 200.59 \, \text{g/mol} + 253.80 \, \text{g/mol} = 454.39 \, \text{g/mol}
\]

2. **Convert the mass of HgI2 to moles**:
We have 0.0118 g of HgI2.
\[
\text{Moles of HgI2} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} = \frac{0.0118 \, \text{g}}{454.39 \, \text{g/mol}}
\]
\[
\approx 2.59 \times 10^{-5} \, \text{mol}
\]

3. **Calculate the concentration (in mol/L)**:
The volume of the solution is 250 mL, which is 0.250 L.
\[
\text{Concentration (C)} = \frac{\text{moles of solute}}{\text{volume of solution in L}} = \frac{2.59 \times 10^{-5} \, \text{mol}}{0.250 \, \text{L}}
\]
\[
= 1.036 \times 10^{-4} \, \text{mol/L}
\]

4. **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 \( \boxed{1.04 \times 10^{-4}} \, \text{mol/L} \).