Questions: Stoichiometry Calculating molarity using solute moles A chemist prepares a solution of calcium bromide (CaBr2) by measuring out 1.4 x 10^2 μmol of calcium bromide into a 450 mL volumetric flask and filling the flask to the mark with water. Calculate the concentration in mmol / L of the chemist's calcium bromide solution. Round your answer to 2 significant digits.

Solution Steps

First, we need to convert the amount of calcium bromide from micromoles ($\mu \mathrm{mol}$) to millimoles ($\mathrm{mmol}$).

\[ 1.4 \times 10^{2} \, \mu \mathrm{mol} = 1.4 \times 10^{2} \times 10^{-3} \, \mathrm{mmol} = 0.14 \, \mathrm{mmol} \]

Next, we convert the volume of the solution from milliliters (mL) to liters (L).

\[ 450 \, \mathrm{mL} = 450 \times 10^{-3} \, \mathrm{L} = 0.450 \, \mathrm{L} \]

Now, we calculate the molarity (concentration) of the solution in $\mathrm{mmol/L}$ using the formula:

\[ \text{Molarity} = \frac{\text{Amount of solute in mmol}}{\text{Volume of solution in L}} \]

Substituting the values:

\[ \text{Molarity} = \frac{0.14 \, \mathrm{mmol}}{0.450 \, \mathrm{L}} = 0.3111 \, \mathrm{mmol/L} \]

Finally, we round the result to 2 significant digits.

\[ 0.3111 \, \mathrm{mmol/L} \approx 0.31 \, \mathrm{mmol/L} \]

Final Answer