The molar concentration of the solution you have is much higher than your target concentration, so yes, of course you can dilute in ethanol to whatever concentration is convenient for you, as long as you store it at -20C. Avoid diluting in water for long term storage.

Carlo

Dear Sir. Concerning your issue about how to prepare aliquots of 100 micro molar (uM) of EPA. Now consider the followings:

You have:

      10 mg of EPA in 100 μL ethanol

that is equivalent to:  10 000 μg  in 100 μL ethanol

multiply by 10 to get:    100 000 μg in 1000 μL ethanol

that is equivalent to:     100 000 μg in 1 mL ethanol

that is equivalent to:     100 000 000 μg in 1 L ethanol

therefore, the original concentration is 100 000 000 μM

so  in order to prepare 100 micro molar (uM) of EPA, use dilution law: V1 M1 = V2 M2.

You  have to take 1 μL from original solution and complete to 1 L by ethanol.

Hello Ravjibhai:

According to the information you have given, the original concentration of EPA is in mg/100uL, where as your targeted concentration is in umol/L. Your calculation should factor in the molar mass of EPA (302.5)

.

Concentration in your initial solution (in uM) can be calculated as follows::

10 mg / 100 uL   =   0.010 g / 0.1 mL  =  0.100 g / 1 mL  =  100 g / 1 L

=  100/302.5 moles/L   =   0.3306 moles/L  = 330600 uM

Now, you can use V1*M1 = V2*M2:

Say, you want to prepare 1 L of 100uM  solution:

Then, V2 = 1L;  M2 = 100 uM;  M1 = 330600 uM

The volume of the aliquot that should be used for dilution, V1

would be:

V1  =  (V2*M2)/M1    =  (1L * 100uM) / (330600uM)

=  0.0003025 L  =  0.3025 mL  =  302.5 uL

Therefore, if you take 302.5 uL of your starting solution and dilute it to 1 litre, you should get 1 litre of 100 uM EPA.  You may change the initial and final volumes proportionately as required (keeping the same dilution proportion).  For example, you may withdraw  10 uL and dilute it to (1/302.5)*10 = 0.03306 L = 33.06 mL.

You can achieve better precision and more convenience if you do the final dilution by gravimetry.  Assuming your diluent is ethanol (density 0.789 g/mL) and neglecting any effect on density by small amounts of EPA, 33.06 mL  =  33.06*0.789 g  = 26.08 g of solution.

Therefore, if you withdraw 10 uL of your original stock solution and dilute it with ethanol on a weighing scale to 26.08 g you will get the required solution quite accurately and much more easily than measuring exact volumes.

Please double check the calculations.

Best wishes,

Nimal