Use this..its that simple...!!

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

I guess your 37%HCL has 12 mol/L?

--> 1kg x 0,37=370gHCL

370g/36,5g/mol=10,137mol/kg

1l=1,185kg

10,137 x 1,185kg/l =12,0mol/l

(x/ml(12mol HCL)*12mol/L)/100ml=0,025mol/L

x=

I hope this helps?!

Cheers, Nadine

Use this..its that simple...!!

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

I think you mean 0.025 M and this is the answer: if your final volume is for example 100 ml, you have to take 208 microliters of stock solution and bring it to 100ml with water.

yes, this is basic biochemistry! sad sigma company and others no longer states the biophysical/biochemical data on their flasks....I have contacted sigma and they answer that there is no free space on the label to write the data....beurk :-)

The concentration of HCl is 12M so by using N1V1 = N2V2. you can prepare.

Firstly, calculate the molar using the following equation M = 10XdensityXpercentage(37 in your case)/Molecular Weight. Then calculate the required molarity by dilution.

Hi Dinesh,

My calculations:

100g of conc. HCl has 37g of HCl

density is 1.2 g/ml

volume = 100g / 1.2(g/ml) = 83.3 ml = 0.0833 L

n = m/MW n = 37g /34.5(g/mol) = 1.07 moles

[HCl] = 1.07 moles / 0.0833 L = 12.8 M!

V1 x C1 = V2 x C2

12.8M x Xml = 0.025M x 100ml

X = 0.025M x 100ml / 12.8M

X = 0.2 ml of conc.HCl bring to 100 ml with ddH2O

Hope it helps! Good luck!

Dear Dinesh,

your stock solution of HCl should be 12.18 M, considering that:

- the density is 1.2 g/mL

- formula weight is 36.46 g/mol

- the concentration is 37% w/w.

Ergo...

in terms of percentage, every 100g of stock solution contain 37g HCl

volume= 100g/1.2g/ml=83.3mL ---> 0.0833 L

n moles= mass/MW = 37g/36.46g/mol = 1.0148 mol

M= mol/L = 1.0148 moles/0.0833L = 12.18 M

To make a 0.025 M solution:

Ci*Vi=Cf*Vf ---> 12.18M*X ml = 0.025*100mL --> 0.205mL

Add 0.205 mL of your stock solution to 100 mL deionized water.

Success!

dear dinesh,.as per davide ,the stock will be12.18 M. use simple formula;V1 *N1 =V2* N2

mole = mass / molucular mass

molucular mass of HCl = 36.5 gm/mol

mass = 0.025*36.5

we need the density

interesting ...maths

1- see the 37% HCL botol ; whats the vol.

2-let say its total vol of HCL= 2.5L

3- to get the concentration of pure HCL;

find its mass from HCL density & volume.

4-Mol of pure HCL = mass/MW & Concentration of pure = mol/volume = a

5-since the HCL is 37% ;0.37x concentration of pure HCL =b

6- finally to get 0.025mol ....use M1 V1= M2 V2

where M1=b , V1= ? & M2=0.025mol/L , V2=1L

7-the V1=the amount need to be taken from 37% HCL

the simple equation:required normality *required volume*M.wt (or) Eq.wt /1000= X g.

where: X = weight of the solute

Simply put. 37% SOLUTION means it is containing 37g in 1000ml which in turn means 1 Molar solution app. Hence take 25ml (0,925g) and dilute further to 1000ml to get 0.025molar solution.

Yes. you are right and we may have to take 2.5g of solution and dilute to 1000ml should be approximately 0.025 molar solution.

Chinnapillai, are we talking about approximate concentration here or having a more accurate concentration? Do you know the density of 37% HCl is 1.19g/l and not exactly the same as water? So 2.5g of HCl should give a volume of 2.5/1.19=2.1ml. I also wonder why you assume 37% HCl should have a molar concentration of 1.00M. Concentration of stock acid solutions should be given on the container label and I pressume 37% HCl has a Conc of ~12.0M or can be calculated using the simple equation; [(% x d) / MW] x 10 = Molarity .

Dinesh, your question should be asking about how much of 37% HCl is need to prepare a certain volume and concentration (Molar not moles because Molar means moles per litre) of solution. If however you meant to say 0.025M and assuming you want to prepare 1L solution then the volume of 37% HCl needed should be=(1000 X 0.025X 36.5)/(37X1.19X10)=2.07 mL diluted to 1 L using DI water.

MOHSEN AHMADI answer is not ciear. either V / V OR w / v shouid be mentioned.

37% HCL has approhimatelly molarity of 11.64 mol/dm3.

To prepare 0.025 M solution you should take:

(0.025x1000)/11.64=2.15 ml of 37% HCV and diluted to 1000 ml with water.

Best regards

Dear Sir, the following method is quoted from the BP which more usefull in daily routine practice and saving time for calculations. The idea is to multiply the factor below (85) by the deisred molarity concentration which will give the volume requirred to be taken from the concentrated HCl acid and diluted up to 1000 mL.

Hydrochloric Acid   HCl = 36.46

Where no molarity is indicated use analytical reagent grade of commerce with a relative density of about 1.18, containing not less than 35% w/w and not more than 38% w/w of HCl and about 11.5M in strength.

Solutions of molarity xM should be prepared by diluting 85x ml of hydrochloric acid to 1000 ml with water .

the spicific gravity of 37% HCl solution is 1.175 . The molarity Of HCl equal (37x 1.175x1000)/(100x36.5) =11.91

MV=M1V1

11.91xV= 0.025x1000

V=2.1mL

transfer 2.1 mLof HCL (37%) to iL volumetric flask then complete to1L with distilled water

Hi

First check the specific gravity of the commercial stock. It is probably 1.175. Since the gravity of water is approx 1; this value equates to approx 1.175 gms/ml i.e. 1175 gms/lit

Then calculate the molarity of the commercial solution - Molarity = moles/lit = Mass (gms)/Mol wt/Vol (lit) i.e. 1175/36.5 = 32 M

Thus the commercial stock is supplied at 32 M

A 37% solution would have 37 ml diluted to 100 ml in water. Thus it would be 0.37 fold dilution and therefore 32 X 0.37 = 11.8 M

Now if you want a 0.025 M solution (molar not moles) in a volume of say 100 ml, use the following formula, N1V1=N2V2

0.025 M X 100 ml = 11.8 M X (x) ml

Thus x = 0.025X100/11.8 = 0.211 ml

Take 0.211 ml of your 37% HCl solution and make up to 100 ml with water

Hope that helps

1) Add 2.053 mL of (37%) HCL solution to 250 mL deionized water.

2) Complete the final volume you want by adding deionized water.  

I recommend Muhammad's answer

Best Regards

this is very very useful calculater for HCl solution please just click the link.

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

We Know,

Molarity M=1000W*Purity/mV

Here, M= Molarity, W=Density*Volume, m=Atomic mass, V=Volume

So, M=1000*Density*Volume*37%/Atomic mass of HCl*Volume

M=1000*1.175*37/36.5*100

M=11.8M

Next, V1S1=V2S2

Here, V1=Required Volume

S1= Stock Concentration

V2=Final Volum

S2= Final Concentration.......Thanks@@

https://www.periodni.com/preparation_of_solutions.php?scq=H2SO4

Since the concentrated HCl is quite inconvenient and hard to handle it is just impossible to take a small volume properly, this way the direct one-step dilution will not recommended and surely would not result in a correct concentration. The most reliable method is to make a 10x dilution first, then to determine the exact concentration of this stock solution by titrimetry in order calculate the proper volumes for a further 2nd step dilution to approach the correct final result.

Gyula Váradi

why not using weight instead of volume? any reason?

instead of 0.211 ml we can use a balance and add 0.245 g

it is faster and more reliable, isn't it?

Dineshkumar Sengottuvelu

You'd have to confirm the final molarity by titration or similar.

Farzad Latifeh I agree. It's simply not possible to add a volume of 0.211 mL.

Actually, a university student should be able to do this ...