Final answer:
To make a 0.215 M LiCl solution when diluted to 300.0 mL, 107.5 mL of a 0.600 M LiCl solution is required, based on the dilution equation M1V1 = M2V2.
Step-by-step explanation:
The question asks how many milliliters of a 0.600 M LiCl solution are needed to create a 0.215 M solution when diluted to 300.0 mL with water. To solve this, we can use the dilution equation M1V1 = M2V2, where M1 is the initial molarity, V1 is the volume of the initial solution needed, M2 is the final molarity, and V2 is the final volume of the solution.
Plugging in the known values we get: (0.600 M)(V1) = (0.215 M)(300.0 mL). Solving for V1, we divide both sides by 0.600 M, resulting in V1 = (0.215 M)(300.0 mL) / 0.600 M. Performing the calculation, V1 = 107.5 mL. Therefore, 107.5 mL of the 0.600 M LiCl solution is required to prepare the desired 0.215 M solution when diluted to 300.0 mL.