What is the molarity of HCl in a 10% HCl solution (10% by volume)?

What exactly do you mean by 10% by volume HCl solution? In chemistry this means 10 cm³ of HCl gas dissolved in 100 cm³ solution

Or 100 cm³ HCl gas dissolved in 1000 cm³ or 1.0 litre solution

Now the question is - how many moles HCl gas are in 100 cm³ volume

If we work at STP , 1 mol HCl gas has volume = 22.4 L

Therefore in 100 cm³ HCl gas you have 100 cm³ / ( 22.4 L * 1000 cm³/L) = 0.00446 mol HCl gas dissolved in 1.0 L solution.

Molarity = 0.00446 mol /L HCl .

I am sure that this is not the answer that you want - but I have answered the question as best that I can. Trying to understand your question and the exact wording thereof . Can I ask you : Do you know what the expression ( that you have used ) : 10% by volume actually means? Antonio has opted to use a far more sensible 10% by mass and has arrived at an alternative answer. But I have a strict policy : I answer exactly what has been asked. And I would be surprised if your teacher actually used the wording that you have submitted.

As as a final suggestion : I am of the opinion that most logical expression of the concentration is 10%(m/v) That is 10 g of HCl gas dissolved in 100 cm³ solution . That would give a solution molarity of 2.74 M

• 37% HCl means, 37- g of HCl is present in 100- ml of the solution.
• So 1000- ml of the solution will contain = 370- g of HCl.
• So Molarity of 37% HCl SOLUTION= 370/36.5 = 10.14 M
• So 50 x 1 = 10.14 x V
• So V= 50 x 1/10.14= 4.93- ml
• Therefore taking 4.93- ml of 37% of HCl and diluting it to 50- ml with distilled water (ie, adding 45.07- ml of distilled water to 4.93- ml of 37% HCl) we can obtain 50- ml of 1M HCl.

Molarity (M) is the most commonly used term to describe the concentration of a solution. It is equal to the moles of solute divided by the 1 litre of solution.

M/10 HCl = 0.1 M HCl solution = 0.1 moles / 1 L

No. of moles = mass of compound/ molar mass

0.1 = x/ 36.5

x = 36.5 X 0.1 = 3.65g of HCl (dissolved in 1 L or 1000 ml of solution)

By volume???????

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10% by mass

D = 1.0456 g/mL

m =%CDV = 0.1 x 1.0456 x 1000 mL= 104.56

mol?

104.56/36.5 = 2.87 mol

2.87 mol / 1 L = 2.87 M

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)

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… :-)

The volume can be anything that you wish . You can have 1 cm³ of 1.0 M HCl solution - you can have 1 dm³ of 1.0 M HCl solution and you can have 1 m³ of 1.0 M HCl solution

The volume of 1.0 M HCl solution that contains 1 mol HCl is 1 dm³

If you would like to clarify what you actually want to know - then do so and you will get further help.

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

You require the density of this solution

From my data : density of 37.2%m/m solution HCl = 1.185 g/cm³

Take 1 litre of this solution

This has mass = 1,185 g

Mass of HCl dissolved = 37.2/100 * 1,185 g = 440.8 g

Molar mass HCl = 36.47 g/mol

mol HCl in 440.8 g = 440.8g/ 36.47 g/mol = 12.09 mol

You have 12.09 mol HCl in 1.0 L solution

Molarity = 12.1 M ( 3 significant digits)