N means normality, N=C/eq.

Folin ciocalteu consists of H3PMo12O40 and H3PW12O40, which are used for staining of phenolic chemicals and biologic subunits. Normality of this compound depends on the H+ atoms present in the solution. Since both these subunits have 3 H+ ions, eq would be 1/3 (because the solution consists of both of them). Thus, the C in the equation becomes: C=N*eq=2/3M

But the information above has only informative property, since regardless the normality/molality/molarity of the solution, 10% solution would have [0.1*volume] grams of the solute in solution.

Folin ciocalteu solution has a density of 1.24 g/cm3 (from: http://www.merckmillipore.com/turkey/food-analytics/folin-ciocalteu-s-phenol-reagent/MDA_CHEM-109001/english/p_llWb.s1O65sAAAEhFBgsgn0Y)

Then, from 1 liters of folin ciocalteu solution, 2/3 moles of solute can be obtained, having mass of m=d*V=1.24g/mL*1000mL=1240g of solution. 100 g of this solution would be V=m/d=100g/(1.24g/mL)=80mL of folin ciocalteu solution per 1000mL of beaker.

I hope the calculations are clear.

Wish you luck.

N means normality, N=C/eq.

Folin ciocalteu consists of H3PMo12O40 and H3PW12O40, which are used for staining of phenolic chemicals and biologic subunits. Normality of this compound depends on the H+ atoms present in the solution. Since both these subunits have 3 H+ ions, eq would be 1/3 (because the solution consists of both of them). Thus, the C in the equation becomes: C=N*eq=2/3M

But the information above has only informative property, since regardless the normality/molality/molarity of the solution, 10% solution would have [0.1*volume] grams of the solute in solution.

Folin ciocalteu solution has a density of 1.24 g/cm3 (from: http://www.merckmillipore.com/turkey/food-analytics/folin-ciocalteu-s-phenol-reagent/MDA_CHEM-109001/english/p_llWb.s1O65sAAAEhFBgsgn0Y)

Then, from 1 liters of folin ciocalteu solution, 2/3 moles of solute can be obtained, having mass of m=d*V=1.24g/mL*1000mL=1240g of solution. 100 g of this solution would be V=m/d=100g/(1.24g/mL)=80mL of folin ciocalteu solution per 1000mL of beaker.

I hope the calculations are clear.

Wish you luck.

Thank you so much

How to change 1.9-2.1 N FC reagent to 1 N FC reagent