You typically start with concentrated HCl ~ 12M. It’s important to standardise the 0.2M HCl solution you make as concentrated HCl is made by injecting HCl gas which is not always exactly 12M. You may get the 12M HCl as 37% (w/w).
How do I convert 37% (w/w) HCl into molarity?
% refers to the solution concentration in percentage and (w/w) refers to the solution and solute amount given in grams (% by weight). So a 37% (w/w) solution of HCl contains 37g of HCl per 100g of solution.
The density of 37% (w/w) HCl is 1.2g/ml at 25℃. This means that the weight of 1ml of hydrochloric acid is 1.2g at 25℃.
Molarity refers to the number of moles of the solute present in 1litre of solution. To calculate the molarity of 37% (w/w) HCl you need to know how many moles of hydrochloric acid are present in 1L of solution.
The molecular weight of HCl is 36.458 g/mol: therefore 1mole of HCl is equal to 36.458g of Hydrogen Chloride.
density = mass / volume
or volume = mass / density
The volume of 100g of Hydrochloric acid = 100g / 1.2 g/ml = 83.33ml
Now we calculate how many grams of HCl are present in 100ml of HCl
83.33ml of 37% (w/w) HCl contains 37g of HCl.
100ml of 37% (w/w) HCl contains 44.40g of HCl.
Now calculate the number of moles of HCl in 44.40g of HCl.
moles = (mass)(mr) = (44.40g)(36.458mol/g) = 1.218 moles of HCl
C = mol/v = 1.218mol / 0.1L = 12.178M
Therefore we can say that 100ml of 37% (w/w) HCl contains 12.178M of HCL
How to prepare [math]frac{2}{10}[/math]N HCl or 0.2N HCl in 100ml of volumetric flask?
If you know the Molarity of an acid or base solution, you can easily convert it to Normality by multiplying Molarity by the number of hydrogen (or hydroxide) ions in the acid (or base). Therefore since only one proton fully dissociates from HCl in an aqueous solution a 12.127M solution of HCl has a normality of 12.127N.
[math]N_1V_1[/math] = [math]N_2V_2[/math]
(12.127N)(x) = (0.2N)(100ml)
x = (0.2N)(100ml)/12.127N
x = 1.6492ml of HCl
Therefore you must add 1.6492ml of 12.127M (or 37% w/w) HCl to 98.3508ml of distilled water to obtain a 0.2N or 0.2M HCl solution.
You typically start with concentrated HCl ~ 12M. It’s important to standardise the 0.2M HCl solution you make as concentrated HCl is made by injecting HCl gas which is not always exactly 12M. You may get the 12M HCl as 37% (w/w).
How do I convert 37% (w/w) HCl into molarity?
% refers to the solution concentration in percentage and (w/w) refers to the solution and solute amount given in grams (% by weight). So a 37% (w/w) solution of HCl contains 37g of HCl per 100g of solution.
The density of 37% (w/w) HCl is 1.2g/ml at 25℃. This means that the weight of 1ml of hydrochloric acid is 1.2g at 25℃.
Molarity refers to the number of moles of the solute present in 1litre of solution. To calculate the molarity of 37% (w/w) HCl you need to know how many moles of hydrochloric acid are present in 1L of solution.
The molecular weight of HCl is 36.458 g/mol: therefore 1mole of HCl is equal to 36.458g of Hydrogen Chloride.
density = mass / volume
or volume = mass / density
The volume of 100g of Hydrochloric acid = 100g / 1.2 g/ml = 83.33ml
Now we calculate how many grams of HCl are present in 100ml of HCl
83.33ml of 37% (w/w) HCl contains 37g of HCl.
100ml of 37% (w/w) HCl contains 44.40g of HCl.
Now calculate the number of moles of HCl in 44.40g of HCl.
moles = (mass)(mr) = (44.40g)(36.458mol/g) = 1.218 moles of HCl
C = mol/v = 1.218mol / 0.1L = 12.178M
Therefore we can say that 100ml of 37% (w/w) HCl contains 12.178M of HCL
How to prepare [math]frac{2}{10}[/math]N HCl or 0.2N HCl in 100ml of volumetric flask?
If you know the Molarity of an acid or base solution, you can easily convert it to Normality by multiplying Molarity by the number of hydrogen (or hydroxide) ions in the acid (or base). Therefore since only one proton fully dissociates from HCl in an aqueous solution a 12.127M solution of HCl has a normality of 12.127N.
[math]N_1V_1[/math] = [math]N_2V_2[/math]
(12.127N)(x) = (0.2N)(100ml)
x = (0.2N)(100ml)/12.127N
x = 1.6492ml of HCl
Therefore you must add 1.6492ml of 12.127M (or 37% w/w) HCl to 98.3508ml of distilled water to obtain a 0.2N or 0.2M HCl solution.
You typically start with concentrated HCl ~ 12M. It’s important to standardise the 0.2M HCl solution you make as concentrated HCl is made by injecting HCl gas which is not always exactly 12M. You may get the 12M HCl as 37% (w/w).
How do I convert 37% (w/w) HCl into molarity?
density = mass / volume
or volume = mass / density
The volume of 100g of Hydrochloric acid = 100g / 1.2 g/ml = 83.33ml
83.33ml of 37% (w/w) HCl contains 37g of HCl.
100ml of 37% (w/w) HCl contains 44.40g of HCl.
moles = (mass)(mr) = (44.40g)(36.458mol/g) = 1.218 moles of HCl
C = mol/v = 1.218mol / 0.1L = 12.178M
Therefore we can say that 100ml of 37% (w/w) HCl contains 12.178M of HCL
How to prepare [math]frac{2}{10}[/math]N HCl or 0.2N HCl in 100ml of volumetric flask?
If you know the Molarity of an acid or base solution, you can easily convert it to Normality by multiplying Molarity by the number of hydrogen (or hydroxide) ions in the acid (or base). Therefore since only one proton fully dissociates from HCl in an aqueous solution a 12.127M solution of HCl has a normality of 12.127N.
[math]N_1V_1[/math] = [math]N_2V_2[/math]
(12.127N)(x) = (0.2N)(100ml)
x = (0.2N)(100ml)/12.127N
x = 1.6492ml of HCl
Therefore you must add 1.6492ml of 12.127M (or 37% w/w) HCl to 98.3508ml of distilled water to obtain a 0.2N or 0.2M HCl solution.
You typically start with concentrated HCl ~ 12M. It’s important to standardise the 0.2M HCl solution you make as concentrated HCl is made by injecting HCl gas which is not always exactly 12M. You may get the 12M HCl as 37% (w/w).
How do I convert 37% (w/w) HCl into molarity?
density = mass / volume
or volume = mass / density
The volume of 100g of Hydrochloric acid = 100g / 1.2 g/ml = 83.33ml
83.33ml of 37% (w/w) HCl contains 37g of HCl.
100ml of 37% (w/w) HCl contains 44.40g of HCl.
moles = (mass)(mr) = (44.40g)(36.458mol/g) = 1.218 moles of HCl
C = mol/v = 1.218mol / 0.1L = 12.178M
Therefore we can say that 100ml of 37% (w/w) HCl contains 12.178M of HCL
How to prepare [math]frac{2}{10}[/math]N HCl or 0.2N HCl in 100ml of volumetric flask?
If you know the Molarity of an acid or base solution, you can easily convert it to Normality by multiplying Molarity by the number of hydrogen (or hydroxide) ions in the acid (or base). Therefore since only one proton fully dissociates from HCl in an aqueous solution a 12.127M solution of HCl has a normality of 12.127N.
[math]N_1V_1[/math] = [math]N_2V_2[/math]
(12.127N)(x) = (0.2N)(100ml)
x = (0.2N)(100ml)/12.127N
x = 1.6492ml of HCl
Therefore you must add 1.6492ml of 12.127M (or 37% w/w) HCl to 98.3508ml of distilled water to obtain a 0.2N or 0.2M HCl solution.
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