Questions: A chemist prepares a solution of iron(III) bromide (FeBr3) by weighing out 180.8 of iron(III) bromide into a 300. miL volumetric flask and filling the flask to the mark with water. Calculate the concentration in g / dL of the chemist's iron(III) bromide solution. Be sure your answer has the correct number of significant digits.

Solution Steps

First, we need to convert the volume of the solution from milliliters (mL) to deciliters (dL).

\[ 300 \, \text{mL} = 300 \, \text{mL} \times \frac{1 \, \text{dL}}{100 \, \text{mL}} = 3.00 \, \text{dL} \]

Next, we calculate the concentration of iron(III) bromide in grams per deciliter (g/dL). The mass of iron(III) bromide is given as 180.8 grams.

\[ \text{Concentration} = \frac{\text{mass of solute}}{\text{volume of solution}} = \frac{180.8 \, \text{g}}{3.00 \, \text{dL}} \]

Now, we perform the division to find the concentration.

\[ \text{Concentration} = \frac{180.8 \, \text{g}}{3.00 \, \text{dL}} = 60.27 \, \text{g/dL} \]

Final Answer