culate the volume in milliliters of a \( 1.2 \mathrm{~mol} / \mathrm{L} \) potassium iodide solution that contains \( 150 . \mathrm{g} \) of potassium iodide (KI). Be sure your answer has the correct nber of significant digits.

To calculate the volume of a \( 1.2 \, \text{mol/L} \) potassium iodide (KI) solution that contains \( 150 \, \text{g} \) of KI, we first need to determine the number of moles of KI in \( 150 \, \text{g} \). 1. **Calculate the molar mass of KI:** - Potassium (K) has an atomic mass of approximately \( 39.10 \, \text{g/mol} \). - Iodine (I) has an atomic mass of approximately \( 126.90 \, \text{g/mol} \). - Therefore, the molar mass of KI is: \[ \text{Molar mass of KI} = 39.10 \, \text{g/mol} + 126.90 \, \text{g/mol} = 166.00 \, \text{g/mol} \] 2. **Calculate the number of moles of KI:** \[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{150 \, \text{g}}{166.00 \, \text{g/mol}} \approx 0.9036 \, \text{mol} \] 3. **Calculate the volume of the solution using the concentration:** The concentration of the solution is given as \( 1.2 \, \text{mol/L} \). We can use the formula: \[ \text{Volume (L)} = \frac{\text{Number of moles}}{\text{Concentration (mol/L)}} \] Substituting the values: \[ \text{Volume (L)} = \frac{0.9036 \, \text{mol}}{1.2 \, \text{mol/L}} \approx 0.7529 \, \text{L} \] 4. **Convert the volume from liters to milliliters:** \[ \text{Volume (mL)} = 0.7529 \, \text{L} \times 1000 \, \text{mL/L} = 752.9 \, \text{mL} \] 5. **Determine the number of significant figures:** The mass \( 150 \, \text{g} \) has 2 significant figures. Therefore, we should round our final answer to 2 significant figures. Thus, the final volume of the potassium iodide solution is: \[ \boxed{750 \, \text{mL}} \]