How much concentrated solution would you take to prepare 2.80L of 0.475M HCl by mixing with water?

Hydrochloric acid is usually purchased in a concentrated form that is 37.0% HCl by mass and has a density of 1.20g/mL.

Calculate the molarity of the concd HCl.

1.20 g/mL x 1000 mL x 0.37 x (1/36.5) = about 12 M or so but you do it exactly.

Then mL x M = mL x M
mL x 12 M = 2800 mL x 0.475
Solve for mL of the concd HCl solution.

Hey thank you DrBob222! That really helped!

Well, let's crunch some numbers! To figure out how much concentrated solution you would need, we need to consider the moles of HCl required.

First, we need to convert the volume of the solution from liters to milliliters (since the density is given in g/mL). So, 2.80L is equal to 2800mL.

Now, we need to determine the mass of HCl that would be present in the 2.80L solution. Since the concentration is 0.475M, we multiply that by the volume (in liters) to get the moles of HCl. Moles = concentration × volume = 0.475M × 2.80L = 1.33 moles.

To calculate the amount of concentrated solution needed, we need to find out how many grams of HCl are present in 1.33 moles of HCl. The molar mass of HCl is 36.46 g/mol (1 mole of HCl weighs 36.46 grams).

Therefore, the mass of HCl in the solution is 36.46g/mol × 1.33 moles = 48.37 grams.

Since the concentrated solution is 37.0% HCl by mass, we can set up a proportion to determine the total mass of the solution needed:

(48.37g / X) = (37g HCl / 100g solution)

Solving for X, we find that the total mass of the concentrated solution needed is approximately 130.7 grams.

So, to prepare 2.80L of 0.475M HCl, you would need to take approximately 130.7 grams of the concentrated solution and mix it with water.

Now that we have the answer, I think it's time to SEA-L the deal with some HCl jokes! Why did the chemist NEVER enjoy working with hydrochloric acid?

Because it was too acidic and kept telling bad jokes... it just couldn't find the right balance!

To determine how much concentrated solution of HCl is needed to prepare 2.80L of 0.475M HCl, we can use the equation:

M1V1 = M2V2

Where:
M1 = initial concentration of HCl (in this case, the concentrated solution)
V1 = volume of concentrated solution to be used
M2 = final concentration of HCl (0.475M)
V2 = final volume of the solution to be prepared (2.80L)

First, we need to calculate the moles of HCl needed to prepare 2.80L of 0.475M solution:

Moles = Molarity x Volume
Moles = 0.475M x 2.80L
Moles = 1.33 moles HCl

Now, we can calculate the volume of the concentrated HCl solution needed:

Moles = Mass / Molar mass
Mass = Moles x Molar mass

The molar mass of HCl is approximately 36.46 g/mol.

Mass of HCl needed = 1.33 moles x 36.46 g/mol
Mass of HCl needed = 48.48 g

Next, we need to convert the mass of HCl needed to volume using the density of the concentrated solution:

Volume = Mass / Density

Density of the concentrated HCl solution is 1.20 g/mL.

Volume of HCl solution needed = 48.48 g / 1.20 g/mL
Volume of HCl solution needed = 40.4 mL

Therefore, you would need to take approximately 40.4 mL of the concentrated HCl solution to prepare 2.80L of 0.475M HCl by mixing with water.

To calculate the amount of concentrated solution needed to prepare the desired solution, we will use the equation:

M1V1 = M2V2

Where:
M1 = initial concentration of the concentrated solution
V1 = volume of the concentrated solution taken
M2 = final concentration of the diluted solution
V2 = final volume of the diluted solution

First, let's calculate the moles of HCl needed for the desired solution:

moles = M2 * V2
moles = 0.475 M * 2.80 L

Next, we need to calculate the volume of the concentrated solution needed to obtain those moles of HCl. To do this, we need to determine the moles of HCl in the concentrated solution:

mass = density * volume
mass = 1.20 g/mL * V1

moles_HCl = mass_HCl / molar_mass_HCl
moles_HCl = mass_HCl / 36.461 g/mol

Since the concentrated solution is 37.0% HCl by mass, we can calculate the mass of HCl present:

mass_HCl = 0.37 * mass_concentrated_solution

Combining the equations:

0.37 * mass_concentrated_solution / 36.461 g/mol = moles_HCl

Now we can substitute moles_HCl with the previously calculated moles:

0.37 * mass_concentrated_solution / 36.461 g/mol = 0.475 M * 2.80 L

Solving for the mass_concentrated_solution will give us the answer to your question.