Question

How do you prepare a 250 mL of a 2.35 M HF dilution from a 15.0 M solution

Answer 1

To prepare a 250 mL dilution of 2.35 M HF from a 15.0 M solution, you need to take 39.2 mL of the 15.0 M HF solution and dilute it with enough water to make the final volume 250 mL.

Step-by-step explanation:

To prepare a 250 mL dilution of 2.35 M HF from a 15.0 M solution, you can use the formula C1V1 = C2V2. Rearranging the formula to isolate the desired parameter, we have V1 = (C2V2) / C1. Plugging in the values, we get V1 = (2.35 M)(0.250 L) / 15.0 M = 0.0392 L or 39.2 mL. Therefore, you need to take 39.2 mL of the 15.0 M HF solution and dilute it with enough water to make the final volume 250 mL.

Answer 2

In dilution problems involving known concentrations and volumes of the solution, a widely used equation is:

C1V1 = C2V2

Where:
C1 = Concentration of Solution 1
V1 = Volume of Solution 1
C2 = Concentration of Solution 2
V2 = Volume of Solution 2

Substituting the given values gives: (2.35 M)*(250mL) = (15.0 M)*V2

V2 = 39.17 mL

So, you must initially have 39.17 mL of 15 M solution and dilute it to 250 mL to achieve a 2.35 M solution