Folin–Ciocalteu formula I am searching a long time now. I have a question about the Folin–Ciocalteu formula for quantification of polyphenol. How is the $62.5$ in the equation explained? Where does it come from?
This is the protocol [1, p. 38]:

$\pu{5.0 mL}$ of extract were placed in calibrated flask and filled up to $\pu{25 mL}$ by distilled water. $\pu{1.0 mL}$ of Folin-Ciocalteu reagent, $\pu{10.0 mL}$ of distilled water were added to $\pu{2.0 mL}$ of diluted extract (from $\pu{25 mL}$ calibrated flask) and filled up to $\pu{25 mL}$ by sodium carbonate solution ($\pu{290 g/L}$). An absorbance of prepared sample was measured by means of UV spectrophotometer at $\pu{760 nm}$ after $\pu{30 min}$ of incubation in darkness. The same liquid (with pure water instead of plant extract) was used as a blind test. All determinations were performed in triplicate. The content of polyphenols was calculated from the formula:
$$X = 62.5 \cdot \frac{A_1 \cdot m_2}{A_2 \cdot m_1}$$
where $A_1$ - absorbance of pyrogallol solution, $A_2$ - absorbance of investigated samples, $m_1$ - mass of investigated sample [$\pu{g}$], $m_2$ - mass of pyrogallol [$\pu{g}$].

Reference

Modnicki, D.; Balcerek, M. Herba Polonica 2009, 55 (1), 35–42.
Authors mention that pyrogallol (PG), among other standards, has been purchased from Sigma Aldrich US and used as is without further dilution. The amounts of polyphenols were calculated as PG equivalent by external standard method, and utilizing additivity of absorption coefficients the proportion has been established.

$62.5$ is just a dilution factor for the herb extract probe. From the protocol, first time the extract was diluted $5$ times (from $\pu{5 mL}$ to $\pu{25 mL}$), second time – $12.5$ times (from $\pu{2 mL}$ to $\pu{25 mL}$), resulting in total dilution factor of $5 \cdot 12.5 = 62.5$.