If  you found 34 mg/L phenol content from the  calibration curve, and the total volume was 12.5 ml, it means, 12.5 ml x 34 g/L /1000=0.0465 mg  was the total phenol content in the 12. 5 ml solution.  It was diluted from 0.5 ml extract, so this content is 0.0465 mg/0.5 ml.    You have 250 ml solution, thus 500 times more phenol is in the 250 ml extract (from the 2.5 g raw material) than in the 0.5 ml, thus 0.0465x500*40 (factor between 2.5 and 100)= 930 mg/100 g raw material, namely 0.93 %.

Thanks for your help.

but 12.5 solution from procedure, are you sure it must be considered as dilution?

Dear Oksal, based on your text, the dilution is 12.5 ml. I think, this was 1 ml and diluted to 25 ml (it isd the same as 0.5 ml to 12.5 ml), because there is no 12.5 ml measuring bottle, but I could start only from your text.

thank again you are very helpful

It is very easy to calculate the phenolic content. You can check the following links-

1. http://www.sciencedirect.com/science/article/pii/S0974694313002387

2. http://pelagiaresearchlibrary.com/european-journal-of-experimental-biology/vol3-iss1/EJEB-2013-3-1-449-454.pdf