Final answer:
Approximately 4.99 mL of a 0.266 M LiNO3 solution is required to dilute to 4.7 mL of a 0.075 M LiNO3 solution, using the dilution equation M1V1 = M2V2.
Step-by-step explanation:
To determine how many milliliters of a 0.266 M LiNO3 solution are required to make 4.7 mL of a 0.075 M LiNO3 solution, we can use the dilution formula M1V1 = M2V2, where M1 and V1 are the molarity and volume of the concentrated solution, and M2 and V2 are the molarity and volume of the dilute solution.
Using the provided information:
- Initial molarity (M1): 0.266 M
- Final molarity (M2): 0.075 M
- Final volume (V2): 4.7 mL
To find V1, we rearrange the formula to V1 = (M2V2)/M1. Plugging in the values:
V1 = (0.075 M x 4.7 mL) / 0.266 M = 1.3275 mL / 0.266 M ≈ 4.99 mL
Thus, approximately 4.99 mL of the 0.266 M LiNO3 solution is needed to prepare 4.7 mL of a 0.075 M solution.