Final answer:
To prepare 35 mL of a 0.39M NaCl solution from a 2.9M stock solution, 4.65 mL of the stock solution is required.
Step-by-step explanation:
The question is asking about the preparation of a diluted sodium chloride (NaCl) solution from a more concentrated one. This involves the concept of molarity (M), which is a measure of the concentration of a solution in moles per liter. To find out how many milliliters (mL) of a 2.9M NaCl stock solution are needed to prepare 35 mL of a 0.39M NaCl solution, we can use the dilution formula M1V1 = M2V2, where M1 and V1 are the molarity and volume of the stock solution, and M2 and V2 are the molarity and volume of the desired diluted solution.
Now, plug in the values we have: M1 = 2.9M, M2 = 0.39M, and V2 = 35mL. We solve for V1, which gives us V1 = (M2 * V2) / M1. Thus, V1 = (0.39M * 35mL) / 2.9M = 4.65mL. Therefore, we would need 4.65 mL of the 2.9M NaCl solution to prepare 35 mL of a 0.39M solution.