Final answer:
To prepare 635.0 mL of 1.00 M HCl from 10.5 M HCl, the calculation using the dilution formula yields a required volume of 60.48 mL of 10.5 M HCl.
Step-by-step explanation:
The question asks how many milliliters of 10.5 M HCl are required to prepare 635.0 mL of 1.00 M HCl. To solve this, we use the dilution formula C1V1 = C2V2, where C1 is the concentration of the starting solution, V1 is the volume of the starting solution needed, C2 is the concentration of the final solution, and V2 is the volume of the final solution.
First, plug in the known values: 10.5 M (C1) and 1.00 M (C2) and 635.0 mL (V2). Then, solve for V1 by rearranging the formula: V1 = (C2V2)/C1. Plugging in the values gives us V1 = (1.00 M x 635.0 mL) / 10.5 M. Calculate V1 to get the required volume of the starting solution.
The calculation results in V1 = 60.48 mL approximately. Therefore, 60.48 mL of 10.5 M HCl are needed to prepare 635 mL of 1.00 M HCl.