To calculate the volume in milliliters of a 3.64 mol/L potassium iodide (KI) solution that contains 300 millimoles (mmol) of potassium iodide, follow these detailed steps:
1. Convert millimoles to moles:
Millimoles (mmol) can be converted to moles (mol) by dividing by 1000 because 1 mol = 1000 mmol.
Given that we have 300 mmol of potassium iodide:
[tex]\[
\text{Moles of KI} = \frac{300 \text{ mmol}}{1000} = 0.3 \text{ mol}
\][/tex]
2. Use the definition of molarity to find the volume:
Molarity (M) is defined as the number of moles of solute per liter of solution. Hence, the volume of the solution in liters (L) can be found using the formula:
[tex]\[
\text{Volume (L)} = \frac{\text{Moles of solute}}{\text{Molarity}}
\][/tex]
Given the molarity of the solution is 3.64 mol/L:
[tex]\[
\text{Volume (L)} = \frac{0.3 \text{ mol}}{3.64 \text{ mol/L}} \approx 0.0824176 \text{ L}
\][/tex]
3. Convert the volume to milliliters:
To convert liters to milliliters, multiply the volume by 1000 (since 1 L = 1000 mL):
[tex]\[
\text{Volume (mL)} = 0.0824176 \text{ L} \times 1000 = 82.4176 \text{ mL}
\][/tex]
4. Round the answer to 3 significant digits:
The final step is to round the volume in milliliters to 3 significant digits:
[tex]\[
82.4176 \text{ mL} \approx 82.418 \text{ mL}
\][/tex]
Therefore, the volume of the potassium iodide solution required is 82.418 milliliters.
