Question

A chemist prepares a solution of barium acetate (Ba(C2H3O2)2) by measuring out 28.3 g of barium acetate into a 250. mL. volumetric flask and filling the flask to mark with water
Calculate the concentration in mol/L of the chemist's barium acetate solution. Round your answer to 3 significant digits.

Answer 1

To calculate the concentration in mol/L (molarity) of the barium acetate solution, we need to follow these steps:

1. **Determine the molar mass of barium acetate (Ba(C₂H₃O₂)₂)**:
- Barium (Ba): 137.33 g/mol
- Carbon (C): 12.01 g/mol × 4 = 48.04 g/mol (since there are 4 carbon atoms)
- Hydrogen (H): 1.008 g/mol × 6 = 6.048 g/mol (since there are 6 hydrogen atoms)
- Oxygen (O): 16.00 g/mol × 4 = 64.00 g/mol (since there are 4 oxygen atoms)

Now, add these together to find the molar mass:
\[
\text{Molar mass} = 137.33 \, \text{g/mol} + 48.04 \, \text{g/mol} + 6.048 \, \text{g/mol} + 64.00 \, \text{g/mol} = 255.418 \, \text{g/mol}
\]

2. **Calculate the number of moles of barium acetate in 28.3 g**:
\[
\text{Moles of Ba(C₂H₃O₂)₂} = \frac{\text{mass}}{\text{molar mass}} = \frac{28.3 \, \text{g}}{255.418 \, \text{g/mol}} \approx 0.111 \, \text{mol}
\]

3. **Calculate the concentration in mol/L**:
The volume of the solution is 250 mL, which is equal to 0.250 L. Therefore, the concentration (C) can be calculated as follows:
\[
C = \frac{\text{moles of solute}}{\text{volume of solution in L}} = \frac{0.111 \, \text{mol}}{0.250 \, \text{L}} \approx 0.444 \, \text{mol/L}
\]

Rounding this result to three significant digits, the concentration of the barium acetate solution is:
\[
\boxed{0.444} \, \text{mol/L}
\]