4.53 g in 1000 mL is 40 mM

http://www.sigmaaldrich.com/etc/medialib/docs/Sigma-Aldrich/Product_Information_Sheet/216763pis.Par.0001.File.tmp/216763pis.pdf

mole = weight (g) / Molecular weight, mole = conc (M) x volume (mL) / 1000

1 M = 1000 mM, 0.04M = 40 mM

Regards

Selvan

conc (M) = weight (g) x volume (mL) / molecular wt x 1000

30%of H2O2 is equivalent to 8.82M solution . You can then dilute it down to 40mM (ie 1 in 220.5 dilution).

Interesting, so we have 3 values: 12.7, 8.8 and 9.8 M. Tough situation. As Markus said, your manufacturer is an important point, as the correct percentage and density will vary. Once you know them correctly, above answers will help you in calculating the right amount.

You need to know the density of of the H2O2 solulution to work out the solution.

No need of density to this calculation at all, using a density is indirect calculation

8.82 M is the right one, but do not use density to calculate this, use the below formula

Conc. (M) = weight (g) x 1000 / Molecular weight x required volume

weight(w)= {conc. (C) * mol wt. (M) * volume (V)} / 1000

Hi Markus,

i agree with you. see my way of calculation

% of H2O2 is 30 % w/w

Molecular weight of H2O2 is 34.0147 g/mol

Dilution to 40 mM or 0.04 M (a molarity is one mole of substances in 1000 mL of solution)

Mole = weight (g) / molecular weight

Mole = Conc. (M) x Volume (mL) / 1000

Equate…

Weight (g) / molecular weight = Conc. (M) x Volume (mL) / 1000

Weight (g) = 0.04 M x 1000 mL X 34.0147 / 1000

= 1.360588 g of hydrogen peroxide

1 g of 30 % w/w is contain 0.3 g of H2O2

4.535293 g of 30 % w/w is contains 1.360588 g of H2O2, (i.e. 4.535293 x 100/ 1.360588 = 30 %)

Now the calculation is

4.535 g of given 30 % (w/w) H2O2 should dissolve in 1000 mL of solution (whatever may be the solvent) to give 0.04M solution of H2O2

Thanks

Selvan

For % w/v H2O2 ..........

% of H2O2 is 30 % w/v

Molecular weight of H2O2 is 34.0147 g/mol

Dilution to 40 mM or 0.04 M (a morality is one mole of substances in 1000 mL of solution)

Mole = weight (g) / molecular weight

Mole = Conc. (M) x Volume (mL) / 1000

Equate…

Weight (g) / molecular weight = Conc. (M) x Volume (mL) / 1000

Weight (g) = 0.04 M x 1000 mL X 34.0147 / 1000

= 1.360588 g of hydrogen peroxide

1 mL of 30 % w/v is contain 0.3 g of H2O2

4.535293 mL of 30 % w/v is contains 1.360588 g of H2O2, (i.e. 4.535293 x 100/ 1.360588 = 30 %)

Now the calculation is

4.535 mL of given 30 % (w/v) H2O2 should dissolve in 1000 mL of solution (whatever may be the solvent) to give 0.04M solution of H2O2

Regards

Selvan

MarKus is right, in my H2O2 solution is 30% (W?W0. I didn't account for the density.

@Salvatore, I am taking your idea of introducing a measured parameter density as one of the input. I appreciate you

in the analytical chemist society weighing is highly recommended for solid, liquid, semisolid to increase the accuracy, weighing is more accurate than measuring in volume (do you agree?), then how my first case need to be corrected for introducing a measuring parameter density (source of another error), if one does not know the density, what will be your way of mole calculation & mole concept?.

Again I want to convey my way of calculation is direct mole calculation / mole concept (no mole concept is talking about the density). Introducing a density is indirect; source of error will become two. One is assay of hydrogen peroxide and another is density, as both are measuring parameter, we don’t know the accuracy of their method of analysis.

Regards

Selvan

There is difference between analytical chemist & process chemist, I think you are a process chemist / technologist, am I right salvatore? This time I agree with you as enough discussion were posted. Good bye.

Hey guys ! thanks y'all for such an appropriate discussion. Specially I thanks Dr Marcus for his clear classification. The concentration of stock soln, I got to prepare is Volume/ Volume - So I worked out with Mr. Arvind's calculation. Anyways thanks all for your valuable suggestions/ discussion.

The molarity of 30% H2O2 solution is 9.8 M. To make 40 mM solution, you should dilute 30% solution 245 times, i.e. mixing of 1 mL of 30% H2O2 with 244 ml of H2O will give you 245 mL of 40 mM H2O2.

make your approximate dilution as others have suggested but check the absorbance at 240nm as there may have been some decomposition

extinction coefficient (40 mM-1 cm-1 at 240 nm) for H2O2 .

It is good to know that Karthivasham got the solution finially.

Hey Come on John Smalley ! I can't get you. What you have posted, is it a way to cross check / confirm, whether we have prepared 40mM H202 ? If so can you elaborate. Thanks.

Wow ! that was awesome, this is the first time I got to know this - to re confirm the concentration of the solution prepared. Well after the check, I will get back to you guys soon. Thanks guys !

Thanks to all that's was good

Thanks a lot for every mention

40 mM= 0.04 M H2O2/Lit

....................................................................................................................

0.04 M (H2O2/Lit) * ( 2*16+ 2*1= 34 g/Mol) = 0.04*34= 1.36 g/L of H2O2 (as : C2)

....................................................................................................................

30% of H2O2==> 300 g/Lit of H2O2 (as : C1)

....................................................................................................................

V1= ?

V2 = final volume of diluted or needed volume (usually = 1000 mL)

----------------------------------------------------------------------------------------

Material Balance Eq.

C1 . V1 = C2 . V2

300 g/L * V1 = 1.36 g/L * 1000 mL

___________

V1 = 4.53 mL of H202 30% ==> to 1000 mL H2O

Each 1 mL= 40 mM

Like Markus pointed out, it's odd that the strength of H2O2 is expressed as % (w/w), for, the titration results could be also be converted into and expressed as Molar or %(v/v). I am not sure if it is either just a standard way adopted, or, makes calculations easy for the common folks – like diluting 30% (w/w) to 3% for use as antiseptic!

First you calculate the molarity of H2O2.

Molarity of a liquid= Molarity of a liquid= % Purity/100 x sp. gravity in 1000 g (per L)/M. Wt.

% Purity of H2O2 in Sigma -Aldrich =30 %

s. gravity= 1.11 g/ml i.e. 1.11 x 1000=1110

M. Wt=34.01

Molarity of H2O2=9.79

Then use the formula

M1 V1= M2V2

9.79 V1=0.04 x  10 ml (e.g. required volume)

V1= 0.04 x 10l9.79=0.0408 ml=0.041 ml or 40 micro liler 

Take 41 micro liter of H2O2 and makeup the volume to 10 ml.

The above fornula can be used for the molarity of any liquid.

Just to throw one more opinion into the mix, my calculations for the Sigma stock solution came out to:

30% H2O2 (weight/ weight) = 9.73 M

So I'd agree with my colleagues with answers in the 9.7-9.8 M range.

hey karthi,

Smalley is absolutely right, u need and should check the absorbance at 240 before proceeding, as h2o2 solution do decompose which would change the molar extinction and the molarity too whether its 30% or 3 % or any solution available with you.

      And Sentha, the involvement of density in calculating molraity is way too important, because the solutions are either given as w/v or w/w, and then u definitely need to consider this because somewhere the density by the mnufacturer is written as 1 g/mol or 1.11 g/mol...u check it and u will get to know the difference yourself in the values calculated 

Thank you so much :)

 If I may add, the first thing to know is the expression for this percentage (30%) whether it is in terms of (i) weight of H2O2 by weight of the solution, (ii) weight of H2O2 by volume of solution, and (iii) volume of H2O2 by volume of solution.  The idea is to have the expression of 30% in terms of the units in (ii)  weight of H2O2 (that is easily converted to moles knowing its molecular weight) by volume of solution (since concentration of molarity uses volume of solution in its denominator). However, when the concentration of the solution is expressed in terms of weight as in the first case (i), the density of solution is required in order to convert the weight of the solution to volume of the solution but then when the concentration of the solution is expressed in terms of volume as in the case of (iii), the density of pure H2O2 is required in order to convert the volume of H2O2 to its corresponding weight. Once this 30% is converted to weight of H2O2 by volume of the solution it is now easy for us to convert it to molarity. Knowing that the number of moles of H2O2 in the aliquot of the starting solution (30% H2O2) to use would be just the same as the number of moles in the final solution (40mM H2O2) of known volume we can make use of this equation C1xV1 = C2xV2, where 1 is the starting solution (30%)  and 2 is the final solution (40mM), C in molar concentration, and V volume of the solution in the same units say L, with the aliquot volume of 30% H2O2 (solution 1) to use as the only one unknown.

Hi,

To prepare a solution with specify concentration from the another matter by knowing the amount of purity and density, we use below formula:

C=(10*a*d)/M

a:purity

d:density

and then by using C1*V1=C2*V2

the volume is achieved.

Lot's of incorrect approaches here. The manufacturer should provide a CoA or likewise. The 30% H2O2 I use from Fischer is 9.79M. Use M1V1=M2V2.

Hydrogen peroxide decomposes overtime so the exact concentration of a 30% bottle should be verified for exact use. Classic method is to use iodine titration which will give you the exact concentration of your stock solution. Use that as your M1 and solve the above equation. There are other hydrogen peroxide assays you can use, although the iodine method is by far the most widely accepted method. 

Trying to calculate molarity based on the 30% value is not necessary, as the company will provide you with the molarity. Either way, standard practice is to determine the molaritu via titration, then dilute as needed. As others have mentioned, H2O2 decomposes overtime so relying on manufacturers value or calculating your own using volume/density is not considered accurate enough for published work. Each time you make a dilution, the stock solution concentration should be checked via titration, it's relatively trivial.

Tank you so much

how to prepare  20mM of  H202 

2.26 ml in 1 liter

ANSWER AT

https://books.google.com/books?id=1OcDAAAAMBAJ&pg=PA1313&dq=PERCENT+H2O2+TO+MOLARITY&hl=fr&sa=X&redir_esc=y#v=onepage&q=PERCENT%20H2O2%20TO%20MOLARITY&f=false

I agree with Hooshyar Hassini answer. I used the same calculation when preparing 40 mM of hydrogen peroxide from 30% hydrogen peroxide solution.

I agree with Hooshyar's comment. It can be following worthily. 

How to prepare 96mM Guaiacol (density=1.112g/cm3) for 100 ml

 First you calculate the molarity of H2O2

Molarity of a liquid= % Purity x  sp. gravity in 1000 g.

                                                100 x M. Wt                                      

 H2O2 has

 % purity=30

Sp. gravity=1.11 g/ml

M. wt=34.01

Molarity of H2O2= % Purity (30)x  sp. gravity in 1000 g or g per liter (1.11 x 1000).

                                                100 x M. Wt   (34.01)           

30/100 x 1.11 x1000/34.01=0.3 x 32.637=9.79 M

Molaraty of H2O2=9.79

m1v1=m2v2

9.79 v1=0.04 x 100=4

m1 x 9.79=4

require M=0.04M

m1= 4/9.79=0.408 ml=408 µl

Dissolve 408 µl of H2O2 to give 100 ml total volume.

First you calculate the molarity of Guaiacol

Molarity of a liquid= % Purity x  sp. gravity in 1000 g.

                                                100 x M. Wt                                      

Guaiacol has

 % purity=?

Sp. gravity=1.112 g/ml

M. wt=?

The following is an example of H2O2, similarly you can prepared for Guaiacol.

Molarity of H2O2= % Purity (30)x  sp. gravity in 1000 g or g per liter (1.11 x 1000).

                                                100 x M. Wt   (34.01)           

30/100 x 1.11 x1000/34.01=0.3 x 32.637=9.79 M

Molaraty of H2O2=9.79

m1v1=m2v2

9.79 v1=0.04 x 100=4

m1 x 9.79=4

require M=0.04M

m1= 4/9.79=0.408 ml=408 µl

Dissolve 408 µl of H2O2 to give 100 ml total volume.

Hi KG

Dr. Islam explained this beautifully. Remember when you have purity percentage, you have to multiply with specific gravity to get amount. See the calculation Dr. Islam has mentioned. 

Thanks for complete answer

Followed. Thanks

How to prepare the 15mM H2O2 for making the final volume 100 ml?

As Dr Islam as explained:

1. by considering that you are using 30% H2O2, for the preparation of 15mM dissolve 153 micro-liters of 30% H2O2 in 99.847 ml of distilled water.

HYDROGEN PEROXIDE

c(H2O2) = 15 mmol/L

=======================

To prepare 100 mL of a 15 mmol/L solution of hydrogen peroxide we will need to dilute 0.15293 mL of 30 % H2O2 to a final volume of 100 mL with deionized (distilled) water.

SOURCE:

-------------

Name: Hydrogen peroxide

Formula: H2O2

Formula weight: 34.015 g/mol

CAS Number: 7722-84-1

NFPA: Health 3, Flammability 0, Instability 1, Special OX

CALCULATION:

-------------

The key concept is that the amount of solute in the desired solution must be equal to the amount of solute in the source solution. Remember, the concentration is the amount of a solute divided by the volume of the solution.

Before we make any calculations we have to make sure that we only use one system and one unit of measurement. DO NOT mix measurement systems and units.

Desired solution:

V0 = 100 mL * (1 L)/(1000 mL) = 0.1 L

c0 = 15 mmol/L * (1 mol/L)/(1000 mmol/L) = 0.015 mol/L

Source solutions:

w1 = 30 % = 30 * 1/100 = 0.3

d1 = 1.1121 kg/L * (1 g/L)/(0.001 kg/L) = 1112.1 g/L

--------------------------------

Then, we determine the concentration of the source (stock) solution

c1 = d(H2O2) * w(H2O2) / M(H2O2)

c1 = 1112.1 g/L * 0.3 / 34.0147 g/mol

c1 = 9.80842 mol/L

Since the total amount of solute is the same before and after dilution, the volume of stock solution needed is

V1 = V0 * c0 / c1

V1 = 0.1 L * 0.015 mol/L / 9.80842 mol/L

V1 = 0.00015293 L

To convert the result into a desired unit we will use dimensional analysis again

V(30 % H2O2) = 0.00015293 L * (1000 mL)/(1 L) = 0.15293 mL

PROCEDURE:

-------------

First of all, fill the volumetric flask about halfway with deionized water to avoid violent reactions. NEVER add water to concentrated acid.

Choose a clean pipette of suitable size and transfer the liquid to the volumetric flask. When the whole solution has been drained, touch the tip of the pipette to the side of the volumetric flask to allow the last of the liquid to drain out. DO NOT blow out the remaining solution.

Allow the solution to reach room temperature because a volumetric flask is only accurate at the temperature at which it has been calibrated (usually 20 癈). Very carefully fill the flask to the mark on the neck of the flask, using a dropping pipette to add the last few milliliters of liquid. Mix your solution thoroughly, by inverting the flask and shaking. NEVER hold large volumetric flasks by the neck alone - provide support at the bottom.

Transfer the prepared solution to a clean, dry storage bottle and label it. NEVER store solutions in a volumetric flask.

SAFETY NOTES:

-------------

- When making chemical solutions, always use the appropriate safety equipment.

- As a general rule, always add the more concentrated solution to the less concentrated solution.

- All chemicals that you are unfamiliar with should be treated with extreme care and assumed to be highly flammable and toxic.

use the relation C = 10xPXd/M

application: C = 10x30X1.11/34,01 = 9,79 mol/L

40mM = 0.04M (ex: V=100ml)

C1V1 = C2V2 V2 = 0,04x100/9,79 = 0,4 ml

Hi

34.01g/L=1 Molar(isn't for 100% H2O2) where here Available only 30% H2O2

If 34.01g/l(100%) is 1 Molar and 30% 34.01g/L is hardly 10.2g/L(30%) which would be 0.3 M(Please correct me )

Molarity of 30% Hydrogen peroxide should be less than 100% H2O2

thanks

TRY using molarity calculator, easy and fast

https://www.graphpad.com/quickcalcs/Molarityform.cfm