How many millimeters of a 0.33 solution of CaCl2 are needed if 3.0 moles of CaCl2 are required for an experiment?

To determine how many milliliters of a 0.33 M solution of CaCl2 are needed to obtain 3.0 moles, we can use the relationship between moles, molarity, and volume:

[ \text{Molarity (M)} = \frac{\text{Moles of solute}}{\text{Volume of solution in liters}} ]

Rearranging this equation gives:

[ \text{Volume (L)} = \frac{\text{Moles of solute}}{\text{Molarity (M)}} ]

Substituting the values provided:

• Moles of CaCl2 = 3.0 moles

• Molarity of the solution = 0.33 M

Calculating the volume in liters:

[ \text{Volume (L)} = \frac{3.0 \text{ moles}}{0.33 \text{ M}} \approx 9.09 \text{ L} ]

To convert this volume into milliliters, multiply by 1000 (since 1 L = 1000 mL):

[ \text{Volume (mL)} = 9.09 \text{ L} \times