Questions: A chemist prepares a solution of mercury(II) iodide (HgI2) by weighing out 9.56 mg of mercury(II) iodide into a 350 mL volumetric flask and filling the flask to the mark with water. Calculate the concentration in g / L of the chemist's mercury(II) iodide solution. Round your answer to 3 significant digits.

Solution Steps

First, convert the mass of mercury(II) iodide from milligrams to grams.

Given: \[ 9.56 \, \text{mg} \]

Since \(1 \, \text{mg} = 0.001 \, \text{g}\), we have: \[ 9.56 \, \text{mg} = 9.56 \times 0.001 \, \text{g} = 0.00956 \, \text{g} \]

Next, convert the volume of the solution from milliliters to liters.

Given: \[ 350 \, \text{mL} \]

Since \(1 \, \text{mL} = 0.001 \, \text{L}\), we have: \[ 350 \, \text{mL} = 350 \times 0.001 \, \text{L} = 0.350 \, \text{L} \]

Now, calculate the concentration of the solution in grams per liter (\(\text{g/L}\)) using the formula: \[ \text{Concentration} = \frac{\text{mass of solute (g)}}{\text{volume of solution (L)}} \]

Substitute the values: \[ \text{Concentration} = \frac{0.00956 \, \text{g}}{0.350 \, \text{L}} \]

Perform the division to find the concentration: \[ \text{Concentration} = 0.00956 \, \text{g} / 0.350 \, \text{L} = 0.0273143 \, \text{g/L} \]

Finally, round the concentration to 3 significant digits: \[ 0.0273143 \approx 0.0273 \]

Final Answer