Hello,

If you needed to dilute because of concentration is too high for calibration or instrument you usually follow conservation of mass,

initial concentration x 5 ml + concentration in distilled water x 5 ml = final concentration x (5 + 5 ) ml

Just to show it but concentration in distilled is zero, so it is

initial concentration x 5 ml = final concentration x (10 ) ml

that makes dilution factor 2 when you compare 10 with 5. You need to know final concentration to calculate initial:

initial concentration = final concentration x 2

Rachid Ait Maalem Lahcen Thanks for your valuable comment.

Please Ref to

https://www.omnicalculator.com/chemistry/dilution-factor#dilution-factor-formula

https://www.omnicalculator.com/chemistry/dilution-factor#how-to-calculate-dilution-factor

Dilution:

Dilution is the prosess where a solution is added more of the solvent to decrease the concentration of the solute.

In dilution, the amount of solute does not change, the number of moles are the same before and after dilution.

If subscript "i" represents initial and "f" represents the final values of the quantities involved, we have:

ni = Mi * Vi and nf = Mf * Vf

ni = nf and thus,

Mi * Vi = Mf * Vf [1]

From this we can calculate the final molarity after dilution:

Mf = Mi * Vi / Vf [2]

or we can calculate the needed final volume to reach a wanted final molarity:

Vf = Mi * Vi / Mf [3]

Dilution factor:

Vf/Vi = Mi/Mf. [4]

In your case, let initial concentration be x mol/ ml. After doubling volume, final concentration be x mol/ 2 mL.

So, from equation [4], you can calculate your dilution factor (Na/K) as:

Dilution factor (Vf/Vi) = 2 (Answer).

Hope, this will help.

Mohammad Shahidul Islam Thanks for your valuable explanation

Without knowing the concentration of the solution, it is impossible to answer this question unequivocally. If 5 mL of distilled water is added to 5 mL of a solution, the dilution factor will be approximately 2 only for a highly diluted solution.