To prepare whatever final molarity concentration of a chemical, the first thing is to know the stock concentration. From the final and stock concentration, calculate the dilution ratio and then use this ratio to work out the amount of stock required to the final volume. Salvatore Sotgia has given a very nice explanation.

There are two contradicting results in this discussion: whether the molarity concentration of 30% H2O2 is 8.82M or 9.8M. Both are correct depending on whether 30% refers to w/w or w/v. The correct answer is 30% w/w = 9.8M and 30% w/v = 8.82M. w/w concentration can be converted to w/v concentration by taking into account that the density of 30% H2O2 is not 1, but 1.11. This is exactly the origin of all these arguments. In fact, 9.8M/1.11=8.82M.

Alessandro D'Ulivo caculation is incorrect as H2O2 formula weight is 34 not 30 g/mole.

Salvatore Sotgia's calculation is correct: molarity of a 30% w/v H2O2 solution is 8.82 mol/l.

Ajay Ramesh should clarify whether 30% is w/v or w/w before ascertain 30% H2O2 is 9.8M.

Owoyomi Olanrewaju's caculation is incorrect as H2O2 formula weight is 34 not 30 g/mole

As w/v and w/w are both available in the market, pay attention to this.

ok mam/sir......sorry...........

It a direct ans to my problem..........

but in this case what will be the weight of H2O2

thank you swapnil sir for such a great and full of concept discussion till can i ask a que then at the end what is weight of H2O2 in final 1000ml is it 3.4gm or else???

This way you go, First you've to calculate the strength of H2O2 i.e, strength = (density * assay* 1000)/Molecular Weight *100 .

All these will be given in H2O2 bottle (over the sticker) Now calculating Strength of H2O2 (i.e)

Strength= (1.135*30*1000)/34.01*100

Strength of H2O2 = 10.01 M

If you are required to make out 0.1M H2O2 for 250ml Then you've to make calculation like this, M1*V1= M2V2

M1=10.01 V1= ? M2= 0.1 V2= 250ml substiute these values to that formula the obtained "x"mL is the mL you've to add to make up the solution.

so 2.49 ml of concentrated H2O2 is added to standard flask and then it is make up to the mark in 250 ml Standard measuring Flask.

You could also estimate the molar concentration of your 30% w/v H2O2 through titration with potassium permanganate in presence of H2SO4 as an example. Then prepare your 0.1M solution through dilution law N1V1=N2V2.

dear Sandeep: I think above answers are sufficient for you I just want to add .... in case of H2O2 always use fresh prepare solution don't use as a sock solution since H2O2 will decompose with time and once more for precaution always store in freeze.

hi, I calculated everything, but i cant show you it. and i just tell you m result. first of all you need to prepare measurement flask (100ml) and take 1.6 ml H2O2 hydroperoxide and then add water to this flask until measurement line. it is 0.1m solution.

To salvatore madam in your calculation check is there any unit cancelation. Moleculare weight experesed in grams. Ok fine W/V = g, and g / g is canceled then why the is result expressed in 8.8 Moles. What is mean of your final result in mole. Note the person want 0.1 Mole solution.

:) your calculation is correct when, suppose the solution expressed in g / %. Ex 20 g of solution contains 30 % of solution, but they were not reporting like this to express your answer for % W/V*10. Moreover W/V = density is in it, do you calculate density, How it will be equal W/V = (W/W =1). H2O is universal solvent Volume and weights are equal, is it too for H2O2. Checks your value 8.82 mol/l with your first add up answer 88.2 times dilute the starting solution. I votes up your first clarity answer solution for this problem and I also votes up swapnil for 11.34 ml in I L solution, i.e., 1.134 ml in 100 ml solution.

:) Yes agree your calculation is correct, but it is difficult to understood by direct way :)... The person is doing to lab scale purpose, so i hope the stock solution (your starting solution) in 1 liter is too exceeding. Mr. Swapnil also give the same solution with simple way. The final solution is 0.1 mol = 1.134 ml of H2O2 in 100 ml hope sufficient for his work. Anyway i know you are wright, thanks for your clear statement on the preceding answer. Welcome :)....

No the digids has again problem... K fine :)..

Extermly sorry for not with a discussion, last two days.....I am very happy and satisfied about the discussion over the said issue....I got some new things also a fruitful opinions from respected Rabah Khalil , Apurba Sinhamahapatra, Ganzorig Khurelbaatar, Salvatore Sotgia and Hari Haran sir...... really now i believe researchgate is a best platform to clearify the concepts..

Thanks Mr. Arote, for your comments, and It is a way to platform gathering the people and informations from someone specialized in that area. I personally feel some valuable arguments and discussion lead deep understanding the subjects. You almost welcome and my special thanks to Dr. sotgia.

Thank you very much Mr. Sandeep Arote for your above comment. Never forget that the concentration of H2O2 solution never be constant with time due to high reactivity of H2O2 reagent. So you have always to prepare fresh solution of H2O2 with the best regards and wishes

If you are sure to have 30% w/v solution, it means that you have 30 g H2O2 per 100 mL solution. Then 30/30.4=0.987 moles of H2O2 in 100mL, corresponding to 9.87 M. Therefore you have to dilute the solution by a factor 100, approximately. More exactly you have to dilute 1.013 mL of solution to 100 mL final volume, or 10.13 mL in 1 L final volume.

Go to http://www.sigmaaldrich.com/chemistry/stockroom-reagents/learning-center/technical-library/molarity-calculator.html

30% H2O2 marked on the bottle is the concentration of H2O2 meaning 30g of H2O2 in 100g of 30%-H2O2 solution. The molarity concentration of H2O2 in the solution is 9.97 M. Then one can dilute any volume from this solution depending on the final volume of your solution.

Once the solution is prepared, the molarity of H2O2 can be checked again by Ki-colorimetric method for measuring absorbance of I3- at 352nm.

3I- + H2O2 --> I3- + 2OH-

This kind of measurement can be found on internet

Today I had a confusion regarding the same question between 8.82 M and 9.8 M, finally got it clarified. The molarity of 30% H2O2 is 9.8 M

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

http://chemistry-knowledge.blogspot.de/2009/09/chemistry.html

http://www.scientistsolutions.com/t21512-how+to+calculate+the+molarity+of+30_+hydrogen+peroxide.html

Dear Sandeep, in the future, for you to be able to calculate things like this, it would be good to look up dimensional analysis. All that is is a fancy term for,how to cancel units so that in your case, you end up with moles per liter at 0.1 M. So if the weight/ volume is 30g per 100 ml which is 30% and you divide by 34 g per mole, the gram units cancel and you end up with mole per ml which is easily converted to moles per liter which is molarity, M. Once you search for how to manipulate the units, you can make any kind of calculation you want. I suggest something on the Internet to look for so that you will always be able to make solutions. Take care

To prepare whatever final molarity concentration of a chemical, the first thing is to know the stock concentration. From the final and stock concentration, calculate the dilution ratio and then use this ratio to work out the amount of stock required to the final volume. Salvatore Sotgia has given a very nice explanation.

There are two contradicting results in this discussion: whether the molarity concentration of 30% H2O2 is 8.82M or 9.8M. Both are correct depending on whether 30% refers to w/w or w/v. The correct answer is 30% w/w = 9.8M and 30% w/v = 8.82M. w/w concentration can be converted to w/v concentration by taking into account that the density of 30% H2O2 is not 1, but 1.11. This is exactly the origin of all these arguments. In fact, 9.8M/1.11=8.82M.

Alessandro D'Ulivo caculation is incorrect as H2O2 formula weight is 34 not 30 g/mole.

Salvatore Sotgia's calculation is correct: molarity of a 30% w/v H2O2 solution is 8.82 mol/l.

Ajay Ramesh should clarify whether 30% is w/v or w/w before ascertain 30% H2O2 is 9.8M.

Owoyomi Olanrewaju's caculation is incorrect as H2O2 formula weight is 34 not 30 g/mole

As w/v and w/w are both available in the market, pay attention to this.

Owoyomi sir,

I just want to make some correction in your above answer

It should be (3.401/30) x 100mL= 11.33mL not 1.133mL

thank you

30% H2O2 means 30 g H2O2 per 100 g, and since the density is 1.11, the molarity is 9.8M.

A concentration of 30% w/v hydrogen peroxide implies that for every 100ml of the solution, there is 30g of hygrogen peroxide. Remember that to prepare 1M of a substance, you need the molar weight in 1L i.e for 1M H2O2, you'd need to dissolve 34.01g of the substance in 1L. Therefore, since we have a solution form of the subtance and 30g is contained in 100ml of water, By cross multiplication, 34.01g is contained in 113.37mL. Dissolve that volume in water and you have 1M of the substance. However, for 0.1M, just take 11.34mL of the 30% solution and make up to 1L.

C=n/v ---------- (1)

n = m/m.wt;  v=m/d

Substitute the values of n and v in eqn 1, 

C = (m/m.wt) * (d/m) = d/M.wt ( Units d= g/mL, and M.wt = g/mol)

                                     = mol/mL; we can write 1000 mol/l

So,   (d*1000)/M.wt. This eqution for 100% solution. If bottle contains 30% of H2O2,

(d*1000)/M.wt *0.3

Finally, subitute the values in the bottle

d = 1.11; M.Wt = 34

So, (1.11 *1000)*(0.3)/34  moles/Litre= 9.79 M  ( Stock concentration)

Now you can make 100 mM H2O2 in 1 ml from your stock;

10.2 micro litre in 1 mL is the exact calculation for 0.1 M H2O2 preparation.

If you are not clear, you can contact me

Thanks

Simplest Calculation

(i) Given: 30% of H2O2 (W/V); if density not mentioned

Solution:

 C = n/v = (30/34) moles/100 mL = 0.88 moles/0.1 Lit =

= 8.8 M

This is the stock concentration

(ii) Given: 30% H2O2 (w/w); density = 1.11 g/ml

C=n/v (30/34) moles/ v     ----------- (1)

where ( v = mass in gram/ density = 100 g/ (1.11 g/mL))

             v = 90.09 mL = 0.09 L

Now Substitute the value of  v in equation (1) 

 C = 9.8 M

Take 11.338 ml of 30% (w/v) H2O2 solution and complete them to 1 liter by water.

Here's a summary of the types of calculations for to obtain hydrogen peroxide solutions

The density of this solution depends firstly on the temperature. Even if we skip the effect of temperature, the density of 1.11 as mentioned is for the solution. As you go to pure H2O2, it. goes up to 1.45 g/mL...So, be critical on how and which value of density it is used.

http://www.h2o2.com/technical-library/physical-chemical-properties/physical-properties/default.aspx?pid=11&name=Density-of-H2O2-Solutions

Article The Density of Aqueous Hydrogen Peroxide Solutions1

If you consider density of H2O2 is 1.11. Then molarity of your original H2O2 stock is 9.8M. and if you take 11.2 ml of 9.8M in 1L. gives 0.1M h2O2.

30% w/v solution means for 100 mL of H2O2 solution, it contains 30g of H2O2. To make 0.1 M solution in one litre we need 0.1 moles of H2O2 i.e. 3.401g. It is given that 30g H2O2 is present in 100 mL solution so, 3.401g of H2O2 would be present in 11.3 mL of H2O2 solution. Hence, to make 0.1 M of 1 litre solution we need 11.3 mL of H2O2 solution and then dilute it with 988.7 mL of water.