If the 30% concentration refers to the weight of H2O2 per 100 mL of solution (w/v), the calculation requires knowing its molecular mass, which is 34.0 g/mole.

30% (w/v) is 300 g/L

(300 g/L)/(34 g/mol) = 8.82 mol/L.

1 millimole = 0.001 mol

0.001 mol/(8.82 mol/L)=0.000113 L = 113 µL of 30% (w/v)

For 1 L of a 1 millimolar solution, add 113 µL of 30% (w/v) H2O2 to 1 L of water.

However, according to the Sigma catalog, they supply 30% H2O2 as a (w/w) solution, rather than a (w/v) solution, so the above calculation would be a little bit off, because the density of a 30% H202 solution is greater than the density of water, since the density of H2O2 is 1.45 g/mL.

In fact, a 30% H2O2 listing at Millipore-Sigma gives the density of 30% H2O2 as 1.11 g/mL, so 100 mL of the solution weighs 111 g, and 100 g of solution has a volume of 100/111 = 90.1 mL. (Caveat: this particular listing did not specify w/v or w/w.)

Therefore, the 30% (w/w) solution contains 30 g of H2O2 per 90.1 mL of solution.

(30 g/90.1 mL) = 0.333 g/mL = 333 g/L

(333 g/L)/(34 g/mol) = 9.79 mol/L

For 1 millimole: (0.001 mol)/(9.79 mol/L) = 0.00102 L = 102 µL

For 1 L of a 1 millimolar solution, add 102 µL of 30% (w/w) H2O2 to 1 L of water.

If you can't find out whether you are using a (w/w) or (w/v) solution, at least you know that 1 millimole is contained in either 102 or 113 µL.

Sir, If I used 100microliter 35% H2O2 in 200ml solution for the experiment, that means it has the around 5.5milimole h2O2 concentation? Am I right?

If 30% H2O2 (w/v) has a concentration of 9.79 M, then 35% (w/v) H2O2 would have a concentration of about 11.4 M.

0.1 mL/200 mL x 11.4 M = 0.0057 M = 5.7 mM

For calculating the molarity of H2O2, I always take the OD at 240 nm, after diluting it in the desired/specific ratio (1:10) and then back calculate the concentration. For my H2O2 30% stock, I have never got a concentration of 8M!! So I rely on the real time concentrations.