Until now to execute this recipe, each time i had to count those milligrams/Liter. An enough difficult procedure.

I am posting here this question to help me understand the procedure of creating stock solutions. So the next time i will create that plant substrate, to take a specific volume of that stock solution.

Thank you.


Since it contains agar, you wouldn't be able to make a stock solution with the agar in it; but you can make a stock solution of the other items and add the agar when you need the working solution. Also, since there is a fair amount of sugar in the recipe, it would be best to add that to the working solution as well.

You should be able to make a 100 times concentrate (leaving out the sugar and agar) and then sterilize it in an autoclave. When you need to make a working solution, weigh 5 g of agar and 20 g of sugar, add 10 mL of stock solution, and about 900 mL of water. Boil the solution to dissolve the agar, cool (to about 50°C) and add water (before it sets) to get 1000 mL of solution.

  • $\begingroup$ Yes, i knew that i have to leave agar and sugar for the end. Also i thought to create two stock solutions. One for the elements that are in a concentration under 10mg/L and one for the rest. The thing is how should i achieve that 100 times concentration you referred. Should i follow the method Saroj provided? $\endgroup$ – F.N Sep 12 '14 at 13:22
  • $\begingroup$ That is the easiest way to get the correct amounts. $\endgroup$ – LDC3 Sep 13 '14 at 0:52
  • $\begingroup$ So to close that thread. 100 times concentration means in that i have to create a stock solution of 100M ? $\endgroup$ – F.N Sep 13 '14 at 11:56
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    $\begingroup$ No, because that would be too much. What you want for boric acid is 1000 mg/L instead of 10 mg/L; cobalt chloride is 2.5 mg/L instead of 0.025 mg/L; etc... Multiply the amounts by 100, but keep the volume the same. $\endgroup$ – LDC3 Sep 13 '14 at 13:06

For the easy reference,you prepare a concentrated stock solution and you dilute it according to your requirement by the below given procedure.You prepare a solution by dissolving a known mass of solute (often a solid) into a specific amount of a solvent. One of the most common ways to express the concentration of the solution is M or molarity, which is moles of solute per liter of solution.


Prepare 1 liter of 1.00 M NaCl solution.

First calculate the molar mass of NaCl which is the mass of a mole of Na plus the mass of a mole of Cl or 22.99 + 35.45 = 58.44 g/mol

1) Weigh out 58.44 g NaCl. 2) Place the NaCl in a 1 liter volumetric flask. 3) Add a small volume of distilled, deionized water to dissolve the salt. 4) Fill the flask to the 1 L line.

If a different molarity is required, then multiply that number times the molar mass of NaCl. For example, if you wanted a 0.5 M solution, you would use 0.5 x 58.44 g/mol of NaCl in 1 L of solution or 29.22 g of NaCl. once your stock solution is prepared you can dilute it to a particular volume to make a new, more dilute solution. The key equation here is

M1V1 = M2V2

Remember that number of moles you are working with is constant whether in the stock solution or in the more dilute new solution. Since molarity times volume equals moles, that it what the equation above is talking about. M1V1 is the molarity (moles per liter) times the volume of the stock solution and M2V2 is the molarity times the volume of the new solution. So you are usually given three of the variables in the problem and you simply plug them in and solve for the missing fourth variable.

In terms of actual laboratory practice to make a new solution from a stock solution, first you calculate how many milliliters of the stock solution that you need to make the new solution. Then measure out the required number of mL's of the stock solution in a graduated cylinder and pour into into a new flask. Then add fresh water (or whatever solvent you are working with) to the flask until the level reaches the required number of mL's of the new solution. Then stir the solution to mix evenly. Now you have a new, more dilute solution of a know concentration.

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    $\begingroup$ You didn't look at his recipe so your answer does not help. $\endgroup$ – LDC3 Sep 12 '14 at 0:52
  • $\begingroup$ I have read this stuff in one link. $\endgroup$ – Freddy Sep 12 '14 at 9:06

The best way of creating a stock solution (even more importante if it is quite diluted) is to start from a highly concentrated solution, and thus dilute it in subsequent steps taking aliquots from the previous one. It is very important that you check the final dilution to be sure of its exact concentration. That is the use of the "analysis standard" sets that permits you to guarantee the concentration of your stock solution.


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