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How can I prepare 3n HCL from 35% HCL?
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+ Chemistry
+ Hydrochloric acid
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Parth Chopra
How can I prepare 3n HCL from 35% HCL?
To answer the question, you need to know the Density of 35% HCl, which is 1.2 gm/mL Thus, a liter of 35% HCl would have a mass of 1200 grams. You also need to decide how much 3 N acid you want. Let’s say 250 mL.
Of the 1200 grams of the concentrated solution, 35% would be HCl. So, you have 420 grams of HCl in a liter of solution. The molecular weight of HCl is 36.5. Therefore you have 420/36.5 moles/liter. The molarity of the 35% acid would be 11.51 molar.
Now, since HCl is a strong, monobasic acid, the molarity and the normality are the same. You can use the equation N1V1=N2V2
11.51 x V1 = 3 x 0.250 (please note that we are using liters as the volume unit: 250 mL = 0.250 L. The answer will be given in liters )
x = 0.0651 L which equals 65.1 mL of the 35% acid mixed with 250–65.1 = 184.9 mL of water will result in 250 mL of 3N acid.
To answer the question, you need to know the Density of 35% HCl, which is 1.2 gm/mL Thus, a liter of 35% HCl would have a mass of 1200 grams. You also need to decide how much 3 N acid you want. Let’s say 250 mL.
Of the 1200 grams of the concentrated solution, 35% would be HCl. So, you have 420 grams of HCl in a liter of solution. The molecular weight of HCl is 36.5. Therefore you have 420/36.5 moles/liter. The molarity of the 35% acid would be 11.51 molar.
Now, since HCl is a strong, monobasic acid, the molarity and the normality are the same. You can use the equation N1V1=N2V2
11.51 x V1 = 3 x 0.250 (please note that we are using liters as the volume unit: 250 mL = 0.250 L. The answer will be given in liters )
x = 0.0651 L which equals 65.1 mL of the 35% acid mixed with 250–65.1 = 184.9 mL of water will result in 250 mL of 3N acid.
37% HCl means 37 gms /100mL or 370 gms /litre.That means 370/36.5 M.Now to convert it to 5.0N HCl. Using N1V1 = N2V2; Here N1 =370/36.5N, V1 = ? ( to be calculated); N2 =5.0N , V2 = 1000 mL ( Assumed ) it can be any value you desire ); Or ( 370/36.5)N *V1 = 0.1N *1000 or V1 = 5.0N *1000/(370/36.5)N; or V1 = 5.0*1000*36.5/370; or V1 = 493mL So take 507 mL of D.M water and mix 493 mL of acid in it to get 1000mL of 5.0N HCl.
37% HCl means 37 gms /100mL or 370 gms /litre.That means 370/36.5 M.Now to convert it to 5.0N HCl. Using N1V1 = N2V2; Here N1 =370/36.5N, V1 = ? ( to be calculated); N2 =5.0N , V2 = 1000 mL ( Assumed ) it can be any value you desire ); Or ( 370/36.5)N *V1 = 0.1N *1000 or V1 = 5.0N *1000/(370/36.5)N; or V1 = 5.0*1000*36.5/370; or V1 = 493mL So take 507 mL of D.M water and mix 493 mL of acid in it to get 1000mL of 5.0N HCl.
N is normality. No one uses that any more. For HCl, normality is equivalent to molarity. So 1N is the same as 1 M or 1 mol/liter. Concentrated HCl is about 37% HCl by mass, and is about 12 M, so 32% would be about 10 M. To make 1 M just make a 1/10 dilution: 1 mL 32 % HCl to 9 mL water.
N is normality. No one uses that any more. For HCl, normality is equivalent to molarity. So 1N is the same as 1 M or 1 mol/liter. Concentrated HCl is about 37% HCl by mass, and is about 12 M, so 32% would be about 10 M. To make 1 M just make a 1/10 dilution: 1 mL 32 % HCl to 9 mL water.
10% volume by volume (v/v) is 10ml of conc. HCl diluted to 100ml with water. Sigma Aldrich states that its concentrated HCl is 12.1M. Therefore, a 1 in 10 dilution makes it 1.21M
10% weight by volume (w/v) or weight by weight (w/w), that would mean 10g of HCl dissolved in water and diluted to 100ml with the same solvent.
That would work out to 2.743M (RMM of HCl 36.46g/mol)
Most conc HCl bottles come at 37% assay value, that is the industrial standard. Using the Sigma Aldrich concentration of conc. HCl (found above), a 10% assay value HCl would be around 3.3M
10% volume by volume (v/v) is 10ml of conc. HCl diluted to 100ml with water. Sigma Aldrich states that its concentrated HCl is 12.1M. Therefore, a 1 in 10 dilution makes it 1.21M
10% weight by volume (w/v) or weight by weight (w/w), that would mean 10g of HCl dissolved in water and diluted to 100ml with the same solvent.
That would work out to 2.743M (RMM of HCl 36.46g/mol)
Most conc HCl bottles come at 37% assay value, that is the industrial standard. Using the Sigma Aldrich concentration of conc. HCl (found above), a 10% assay value HCl would be around 3.3M
To answer the question, you need to know the Density of 35% HCl, which is 1.2 gm/mL Thus, a liter of 35% HCl would have a mass of 1200 grams. You also need to decide how much 3 N acid you want. Let’s say 250 mL.
Of the 1200 grams of the concentrated solution, 35% would be HCl. So, you have 420 grams of HCl in a liter of solution. The molecular weight of HCl is 36.5. Therefore you have 420/36.5 moles/liter. The molarity of the 35% acid would be 11.51 molar.
Now, since HCl is a strong, monobasic acid, the molarity and the normality are the same. You can use the equation N1V1=N2V2
11.51 x V1 = 3 x 0.250 (please note that we are using liters as the volume unit: 250 mL = 0.250 L. The answer will be given in liters )
x = 0.0651 L which equals 65.1 mL of the 35% acid mixed with 250–65.1 = 184.9 mL of water will result in 250 mL of 3N acid.
To answer the question, you need to know the Density of 35% HCl, which is 1.2 gm/mL Thus, a liter of 35% HCl would have a mass of 1200 grams. You also need to decide how much 3 N acid you want. Let’s say 250 mL.
Of the 1200 grams of the concentrated solution, 35% would be HCl. So, you have 420 grams of HCl in a liter of solution. The molecular weight of HCl is 36.5. Therefore you have 420/36.5 moles/liter. The molarity of the 35% acid would be 11.51 molar.
Now, since HCl is a strong, monobasic acid, the molarity and the normality are the same. You can use the equation N1V1=N2V2
11.51 x V1 = 3 x 0.250 (please note that we are using liters as the volume unit: 250 mL = 0.250 L. The answer will be given in liters )
x = 0.0651 L which equals 65.1 mL of the 35% acid mixed with 250–65.1 = 184.9 mL of water will result in 250 mL of 3N acid.
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37% HCl means 37 gms /100mL or 370 gms /litre.That means 370/36.5 M.Now to convert it to 5.0N HCl.
Using N1V1 = N2V2;
Here N1 =370/36.5N, V1 = ? ( to be calculated);
N2 =5.0N , V2 = 1000 mL ( Assumed ) it can be any value you desire );
Or ( 370/36.5)N *V1 = 0.1N *1000
or V1 = 5.0N *1000/(370/36.5)N;
or V1 = 5.0*1000*36.5/370;
or V1 = 493mL
So take 507 mL of D.M water and mix 493 mL of acid in it to get 1000mL of 5.0N HCl.
37% HCl means 37 gms /100mL or 370 gms /litre.That means 370/36.5 M.Now to convert it to 5.0N HCl.
Using N1V1 = N2V2;
Here N1 =370/36.5N, V1 = ? ( to be calculated);
N2 =5.0N , V2 = 1000 mL ( Assumed ) it can be any value you desire );
Or ( 370/36.5)N *V1 = 0.1N *1000
or V1 = 5.0N *1000/(370/36.5)N;
or V1 = 5.0*1000*36.5/370;
or V1 = 493mL
So take 507 mL of D.M water and mix 493 mL of acid in it to get 1000mL of 5.0N HCl.
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Hello,
If you have HCl in liquid form then for 100 mL 10% HCl,
Take 10 mL of HCl and mixed with 90 mL of Distilled Water. (Take Water first i.e. in 90 mL of water add 10 mL HCl) it will give you 10% v/v HCl.
If you have solid HCl (powder) then for 100 mL,
Take 10 gm of HCl powder then add 100 mL water. It will give you 10% w/v HCl.
Actually final volume should be 100.(maybe around 99.8 mL water required). But that is not going to affect that much.
Hope this will Help… :-)
Hello,
If you have HCl in liquid form then for 100 mL 10% HCl,
Take 10 mL of HCl and mixed with 90 mL of Distilled Water. (Take Water first i.e. in 90 mL of water add 10 mL HCl) it will give you 10% v/v HCl.
If you have solid HCl (powder) then for 100 mL,
Take 10 gm of HCl powder then add 100 mL water. It will give you 10% w/v HCl.
Actually final volume should be 100.(maybe around 99.8 mL water required). But that is not going to affect that much.
Hope this will Help… :-)
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1
in 1 L solution???
D = 1.174
%C = 35
3 mol/ 1 L
3 x36.5 = 109.5 g
109.5 = 0.35 x 1.174V
V = 266.5 mL
take this volume and dilut it carefully with wáter till a volume of 1 L
in 1 L solution???
D = 1.174
%C = 35
3 mol/ 1 L
3 x36.5 = 109.5 g
109.5 = 0.35 x 1.174V
V = 266.5 mL
take this volume and dilut it carefully with wáter till a volume of 1 L
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1
10% by mass; density = 1.04555 g/mL
Normality????
%C = m/DV x 100
DV = 100 g
1.04555 V = 100
V = 95.64 mL
Normality=
10 g/ 36.5/0.09564 L = 2.87N
V1N1 = V2N2
12.1 (0.09564 -X)= 2.87 x 0.09564
X = 0.073
V1 = 0.09564–0.073 = 22.64 mL
V1 = 22.64 mL
Take this volume from the 12.1 N solution and dilute it (carefully) till a final volume of 95.64 mL
%m/v?????????
10g in 100 mL
10g / 36.5 / 0.1 L =2.74 N
V1N1 = V2N2
12.1 (0.1 -X) = 2.74 x 0.1 L
X = 0.0774 ml
0.1 -0.0774 = 22.6 mL
take this volume from the intial acid and dilute it till 100 mL (carefully)
10% by mass; density = 1.04555 g/mL
Normality????
%C = m/DV x 100
DV = 100 g
1.04555 V = 100
V = 95.64 mL
Normality=
10 g/ 36.5/0.09564 L = 2.87N
V1N1 = V2N2
12.1 (0.09564 -X)= 2.87 x 0.09564
X = 0.073
V1 = 0.09564–0.073 = 22.64 mL
V1 = 22.64 mL
Take this volume from the 12.1 N solution and dilute it (carefully) till a final volume of 95.64 mL
%m/v?????????
10g in 100 mL
10g / 36.5 / 0.1 L =2.74 N
V1N1 = V2N2
12.1 (0.1 -X) = 2.74 x 0.1 L
X = 0.0774 ml
0.1 -0.0774 = 22.6 mL
take this volume from the intial acid and dilute it till 100 mL (carefully)
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1
N is normality. No one uses that any more. For HCl, normality is equivalent to molarity. So 1N is the same as 1 M or 1 mol/liter. Concentrated HCl is about 37% HCl by mass, and is about 12 M, so 32% would be about 10 M. To make 1 M just make a 1/10 dilution: 1 mL 32 % HCl to 9 mL water.
N is normality. No one uses that any more. For HCl, normality is equivalent to molarity. So 1N is the same as 1 M or 1 mol/liter. Concentrated HCl is about 37% HCl by mass, and is about 12 M, so 32% would be about 10 M. To make 1 M just make a 1/10 dilution: 1 mL 32 % HCl to 9 mL water.
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Depends what you mean 10%.
10% volume by volume (v/v) is 10ml of conc. HCl diluted to 100ml with water. Sigma Aldrich states that its concentrated HCl is 12.1M. Therefore, a 1 in 10 dilution makes it 1.21M
10% weight by volume (w/v) or weight by weight (w/w), that would mean 10g of HCl dissolved in water and diluted to 100ml with the same solvent.
That would work out to 2.743M (RMM of HCl 36.46g/mol)
Most conc HCl bottles come at 37% assay value, that is the industrial standard. Using the Sigma Aldrich concentration of conc. HCl (found above), a 10% assay value HCl would be around 3.3M
Depends what you mean 10%.
10% volume by volume (v/v) is 10ml of conc. HCl diluted to 100ml with water. Sigma Aldrich states that its concentrated HCl is 12.1M. Therefore, a 1 in 10 dilution makes it 1.21M
10% weight by volume (w/v) or weight by weight (w/w), that would mean 10g of HCl dissolved in water and diluted to 100ml with the same solvent.
That would work out to 2.743M (RMM of HCl 36.46g/mol)
Most conc HCl bottles come at 37% assay value, that is the industrial standard. Using the Sigma Aldrich concentration of conc. HCl (found above), a 10% assay value HCl would be around 3.3M
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