Dear Soha Ye,

If I understand your question correctly then you wonder how to make a 1 mM solution which sounds as such a low concentration to you that you wonder how you can weigh the salts properly.

Based on:

MW KCl = 74,5513 g/mol

MW MgCl2 = 95,211 g/mol

This would mean 74, 55 mg KCl in 1 L buffer and 95, 21 mg MgCl2 in 1 L buffer respectively. Which is doable by the way.

However more common is it to prepare for example 0.1 M (or 100 mM) solutions of your salts and add for example 1 mL salt solution to 100 (or better 99) mL of your buffer solution. This way your salt solution is diluted in this case a 100 fold.

Preparing a 0.1 M solution of your salt would mean:

7, 4 gram in 1 L (or 0.74 g in 100 mL) KCl and 9,5 gram in 1 L MgCl2 (or 0,95 g in 100 mL).

Guess this is what you mean and answers your question.

The following discussion seemingly can be found of some interest, concerning to your query: https://www.researchgate.net/post/I_tried_to_prepare_a_MgCl2_4M_solution_using_anhydrous_magnesium_chloride_overnight_but_doesnt_dissolve_completely_in_water_Any_ideas