Calculate the volume in milliliters of a 1.34 M barium acetate solution that contains 225. mmol of barium acetate \( \left(\mathrm{Ba}\left(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)_{2}\right) \). Round your answer to 3 significant digits. \( \square \square_{\mathrm{mL}} \)

To find the volume of the 1.34 M barium acetate solution that contains 225 mmol of barium acetate, we can use the formula: \[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{liters of solution}} \] First, convert 225 mmol to moles: \[ 225 \, \text{mmol} = 0.225 \, \text{mol} \] Next, use the molarity to find the volume: \[ 1.34 \, \text{M} = \frac{0.225 \, \text{mol}}{V \, \text{L}} \] Rearranging to solve for \( V \): \[ V = \frac{0.225 \, \text{mol}}{1.34 \, \text{M}} \approx 0.1672 \, \text{L} \] Now convert liters to milliliters: \[ 0.1672 \, \text{L} = 167.2 \, \text{mL} \] Rounding to three significant digits, the volume is: \[ \boxed{167 \, \text{mL}} \]