You can find the density from place where you ordered and/or on the package/bottle of the reagent. In sigma aldrich's site it is given that d=1.03 g/mL

Which means, in 1L of solution you have m/1000mL=1.03g/mL   m=1030g

Now, to get 1mM solution of 50mL you should have 224.3/(1000/(1000/50))=11.2mg

11.2mg will correspond to 1.09x10^-5 of the 1L solution, which will be 10.9 microlitres of solution. Considering the fact the it is %95 solution, you should have 10.9/0.95=11.5microliters taken from stock solution and filled with the solvent to fill 50mL solution.

I hope calculations are clear.

Good luck

Dear Muhammad,

Please read the following:

1M 100% methyljasmonate = 224.3 g/Liter; since you have 95% methyljasmonate you should use 236.1 g/Liter to obtain 1M.

1mM = 236.1 mg/Liter; since you need only 50 mL; 

50/1000 x 236.1 = 11.81 mg; therefore for obtaining 50 mL of 1mM solution you should wegh 11.81 mg of 95% methyljasmonate, dissolved in sufficient quantity of the solvent and complete the volume to 50 mL.

Hoping this will be helpful,

Rafik

Thank you for yours answer

but how will i get 11.81 mg from 95% aqueous solution of methyljasmonate.  

You can find the density from place where you ordered and/or on the package/bottle of the reagent. In sigma aldrich's site it is given that d=1.03 g/mL

Which means, in 1L of solution you have m/1000mL=1.03g/mL   m=1030g

Now, to get 1mM solution of 50mL you should have 224.3/(1000/(1000/50))=11.2mg

11.2mg will correspond to 1.09x10^-5 of the 1L solution, which will be 10.9 microlitres of solution. Considering the fact the it is %95 solution, you should have 10.9/0.95=11.5microliters taken from stock solution and filled with the solvent to fill 50mL solution.

I hope calculations are clear.

Good luck

Dear Elbay,

You did not consider the 95%. Therefore the exact answer will be 11.81 mg as I indicated in my previous answer.

Rafik

Dear Rafik:

At last paragraph I wrote 

"Considering the fact the it is %95 solution, you should have 10.9/0.95=11.5microliters taken from stock solution"

This operation can be done at any stage of calculation, at start or at end. It doesn't change the answer.

And yes, if you would add the density into equation, the result would be:

11.81mg/1.03g/mL (*10^-3) = 11.5 microliters. The same answer that I got.

Good luck

Dear Muhammad,

As Elbay wrote: you should check the density of the product and if is hard to take small volumes you might take 10-fold volume and then make a dilution of 10.

Volume = Weight/Density.

Good luck,

Rafik

Thank you all. 

Kind regards 

There is a much simpler way to figure this, i have been on this thread for ages and Elbay there is an error in your calculations

As it is given as 95%, you can see that is 95g per 100ml, but molarity is based upon litres, so you must * this total by 10 to reconvert to litre scale (950g/L). This can then be divided by the formula weight of pure MeJa 95% stock (224.3g/mol)

(950)*10

------------------------------- = 4.235 M

224.3

So now we know the stock itself has the concentration of 4.235 M it is simple enough to get your required amount

So with a 4.235 M stock, which you want to produce 50ml of 1mM stock, you need to add 11.8 ul of MeJa 95% stock

Hope that helps

[EDIT 27.10.19 : I have edited this post to simplify and have a full written explanation below]

Dear Sir

I could not understand you calculation. could you please expain sir

Ganesan Mahendran I have edited my first comment in the hopes it has simplified the working, and also have attached a full breakdown of the calculations

Excuse the paper wasn't near a PC

Hope it helps