To prepare 0.1M HCl we should be given as
Density of HCl=1.19 g/ml approx.
Purity of HCl=37% w/w approx.
Molecular weight of HCl=36.5 approx.
Now (i)
Convert w/w into w/v of HCl
37% × density = 37/100×1.19= 0.4403 kg/L
Now(ii)
Convert w/v into mol/v
0.4403 kg/L / 36.5 g/mol ×1000= 12M (HCl)
Now(iii)
M1V1=M2V2
12×V1= 0.1 × 1000
V1= (0.1×1000)/12
V1=8.33 ml of HCl will be used to make up volume of water up to 1000 ml for 0.1M HCl.
Exactly how you prepare will depend on what you are starting with. Typically to make a 1 M HCl solution, you will be starting with a stock solution of more concentrated HCl that you will then dilute.
See the Related Questions for complete instructions on how to prepare a solution by diluting a stock solution.
1M HCl Solution*Concentrated HCl. its 37.5%.
*The density of concentrated HCl is 1.189g/ml we will need this number as well, and of course the Atomic Mass of HCl 36.46
* Calculation=((37.5/100)(1000)(1.189))/36.46 = 12.2M
*If HCl concentrated is 12.2 M then to make a 1L solution of 1M HCL
(12.2) x = 1(1)
x = 1(1)/(12.2)
x=0.082ml/ml of water
x = 82mL HCl per liter
0.1M IS THE CONCENTRATION THAT SHOULD BE PREPARED,i.e. C2.WE SHOULD HAVE THE IDEAL VOLUME IN WHICH WE WANT TO PREPARE THE SOLUTION,SAY 2L, IS OUR V2.IF WE ARE TAKING THE HCL FROM A 12M Hcl,C1,THEN WE SHUOLD FIND OUR V1. C1=12M V1=? C2=0.1M V2=2L C1V1=C2V2 12M.V1=0.1M.2L V1=0.2M.L/12M =0.016L =16.6ml ADD 16.6ml OF HCL IN A 2L VOLUMETRIC FLASK AND WITH SOME DISTILLED WATER.SWIRL TO MIX AND ADD DISTILLED WATER UP TO THE MARK OF THE VOLUMETRIC FLASK. MALAPANE L.A For safety purposes you should always add acid to water, especially in the case of an acid like sulfuric acid. When mixing acid and water you should add some of the water to your flask, then add the acid and swirl to mix, then add the balance of the water.
Suppose your question is to prepare 1 mM HCl of 100 ml using 1 M HCl?
Take equation V1N1 = V2N2
V1 is the volume of your 1mM HCl, here V1= 100 ml
N1 is the your desired concentration of HCl. here N1= 1 mM
V2 is the required volume (ml) of 1M HCl to prepare 100 ml of 1mM, here V2
N1 is the concentration of your available stock acid, here N1= 1M
1mM= 0.001 M
Now, V2 = (100ml x 0.001)/1 = 0.1 ml
Therefore, 0.1 ml of 1 M HCl with remaining 99.9 ml of distilled water addition will give 100 ml of 1mM HCl from 1M HCl stock.
To calculate this conncetration we have to use the formula of molarity: M = mol solute / L solution. Now, considering that a stock solution of Hydrochloric Acid (HCl) is calculated to be 12.178 M based on a density of 1.2 g/mL, a formula weight of 36.46 g/mol, and a concentration of 37% w/w.
Therefore, to make 1 L of a 0.1 M solution, we have to slowly add 8.212 mL of the HCl stock solution to 250 mL deionized water, and adjust the final volume of solution to 1000 mL with deionized water.
To prepare a .1 M solution, the ratio would be .1 moles of HCl per liter of water. This is equal to 3.65 grams HCl.
8.3ml HCl from 37% v/v stock bottle is required to prepare 0.1 M HCl solution.(answered by Prof.WASEEM UR RAHMAN KHAN)
2.23 ml of HCl in 100 ml gives 1% solution. 0.23 in 10 ml also gives 1% solution
one mole of HCl in one litre of water, result in 1M HCl
That depends on what you start with. You need 0.01 moles HCl in a total volume of 1 liter (or any multiple of that ratio).
Prepare HCl 1 M by HCl concentration 37 % HCl concentration 37 % have density =1.19 g/ml HCl 1 M use HCl 37 % 82.81 ml make volume with water to 1 liter
dilute your HCl solution to 0.2 M HCl solution and then follow above mentioned link : http://delloyd.50megs.com/moreinfo/buffers2.html
33.3 ml of 3.0 M HCl should be filled up and mixed with water up to 100.0 ml of a 1.0 M HCl solution.
0.2 N HCl solution means 0.2 equivalents of HCl dissolved in 1 litre of water. Normality = Molarity x n-factor => Molarity =Normality/n-factor=0.2/1=0.2 M 0.2 moles should be present in 1 litre of solution. 0.2moles =0.2 x 36.5 = 7.3 grams of HCl =>Dissolve 7.3 grams of HCl in 1 litre if water to get a 0.2N solution.
44.5 ml HCl TAKE AND DILUTE UP TO 1000 ML WATER MAKE A 0.5 M HCl SOLUTION
Prepare HCl 1 M by HCl concentration 37 % HCl concentration 37 % have density =1.19 g/ml HCl 1 M use HCl 37 % 82.81 ml make volume with water to 1 liter
dilute your HCl solution to 0.2 M HCl solution and then follow above mentioned link : http://delloyd.50megs.com/moreinfo/buffers2.html
33.3 ml of 3.0 M HCl should be filled up and mixed with water up to 100.0 ml of a 1.0 M HCl solution.
0.2 N HCl solution means 0.2 equivalents of HCl dissolved in 1 litre of water. Normality = Molarity x n-factor => Molarity =Normality/n-factor=0.2/1=0.2 M 0.2 moles should be present in 1 litre of solution. 0.2moles =0.2 x 36.5 = 7.3 grams of HCl =>Dissolve 7.3 grams of HCl in 1 litre if water to get a 0.2N solution.
44.5 ml HCl TAKE AND DILUTE UP TO 1000 ML WATER MAKE A 0.5 M HCl SOLUTION
you need to make 1000 times dilutions, this could be done in multi-steps: transfer 1 ml 0.1 M HCl into 100 ml volumetric flask and complete volume with water --------(1) from solution (1) transfer 2 ml into 20 ml volumetric falsk and complete volume with water, this is 0.0001 M HCL.
N10 HCl would be 10 M HCl. You might mean N/10 which would be 0.1 M. (N or normal is an old-fashioned way to define chemical concentrations. Most chemists prefer M or molar concentrations.) Just take some higher-concentration HCl and dilute it to get the desired concentration. Use the dilution formula C1V1 = C2V2 to figure out how much high-concentration HCl to use for the volume of dilute HCl you want to make. If anything here doesn't make perfect sense, you probably shouldn't be working unsupervised with chemicals, and if you don't know what HCl is then you definitely shouldn't do whatever it is you are trying to do. 10 M HCl is a caustic liquid that gives off caustic vapors so wear goggles and be careful with it.
The answer is 5 m L sodium bicarbonate, 1 M solution.
(7 mL)(X M HCl) = (27.6 mL)(0.170 M NaOH)7X = 4.692X = 0.7 M HCl==========
It doesn't matter how much you have, HCl (Hydrochloric Acid) has a pH of 1
First get moles HCl, then get the grams HCl.Molarity = moles of solute/Liters of solution3.8 M HCl = X moles/17.2 Liters= 65.36 moles HCl===============so,65.36 moles HCl (36.458 grams/1 mole HCl)= 2382.89 grams HCl======================and,put this amount, 2382.89 grams, 65.36 moles HCl into the 17.2 liters of solution ( water, I suppose ) 2382.89 grams = 2.38 kilograms. I suppose you could weigh out the HCl in a flask of some sort.
The answer is 15,039 g hydrogen chloride (HCl).