how do u prepare a 250-ml of 2.35 M HF dilution from a 15.0 M stock solution?
39.2ml
To prepare a 250 ml of a 2.35 M HF dilution from a 15.0 M stock solution, you will need to use the dilution formula, which is:
C1V1 = C2V2
Where:
C1 = concentration of the stock solution
V1 = volume of the stock solution to be used
C2 = desired concentration of the dilution
V2 = final volume of the dilution
In this case:
C1 = 15.0 M (concentration of the stock solution)
V1 = ?
C2 = 2.35 M (desired concentration of the dilution)
V2 = 250 ml (final volume of the dilution)
Rearranging the formula, we have:
V1 = (C2 * V2) / C1
Now let's calculate V1:
V1 = (2.35 M * 250 ml) / 15.0 M
= 39.17 ml
Therefore, you will need to measure 39.17 ml of the 15.0 M stock solution and then add enough solvent (like water) to make a final volume of 250 ml to prepare a 2.35 M HF dilution.
To prepare a 250 mL of 2.35 M HF dilution from a 15.0 M stock solution, you can use the dilution formula:
(C1)(V1) = (C2)(V2)
Where:
C1 = concentration of the stock solution
V1 = volume of the stock solution
C2 = concentration of the diluted solution
V2 = volume of the diluted solution
In this case:
C1 = 15.0 M (concentration of the stock solution)
V1 = ? (volume of the stock solution)
C2 = 2.35 M (concentration of the diluted solution)
V2 = 250 mL (volume of the diluted solution)
Rearranging the formula:
V1 = (C2)(V2) / C1
Plugging in the values:
V1 = (2.35 M)(250 mL) / 15.0 M
V1 = 39.17 mL
Therefore, you will need to measure 39.17 mL of the 15.0 M HF stock solution and then add enough solvent, such as water, to bring the final volume up to 250 mL to prepare a 2.35 M HF dilution.