Hydrochloric acid is typically purchased in a relatively highly concentrated form and then diluted by adding slowly to water to save money. The concentration of the purchased solution is 37.0% by mass HCl, and this solution has a density of 1.20 g/mL.
What volume of the concentrated HCl solution is the correct amount needed to make 3.7 L of 0.435 M HCl? Please express your answer in mL.
The answer is _________ mL.
Two steps.
1.First, determine the molarity of the 37% HCl.
2. Second, work on the dilutions.
First:
1.20 g/mL x 1,000 mL x 0.37 x (1/36.5) = 12.2 M
Second: Use the dilution formula
mL1 x M1 = mL2 x M2
mL1 x 12.2 = 3,700 mL x 0.435 M
Solve for mL1 which is the amount of the 12.2 M stuff that should be taken.
Well, it seems like someone is taking their chemistry experiments very seriously! Now, let's see how we can clown around with these numbers.
To determine the volume of concentrated HCl solution needed, we first need to find out the amount of HCl in moles.
We have 3.7 L of a 0.435 M HCl solution, which means we have:
0.435 moles/L x 3.7 L = 1.605 moles of HCl
Now, let me pull out my clown calculator to help us calculate the mass of HCl needed to make 1.605 moles of HCl.
Since the concentration of the purchased HCl solution is given in mass percent, we can assume that 37.0 g of the solution contains 100 g of HCl. So:
mass of HCl = (1.605 moles) x (36.5 g/mol of HCl) = 58.5825 g
To find the volume of the concentrated HCl solution, we can use its density. The density of the solution is given as 1.20 g/mL, which means in 1 mL we have 1.20 g of the solution.
So, the volume of the concentrated HCl solution needed can be calculated as:
volume = (mass of HCl) / (density of HCl solution)
= (58.5825 g) / (1.20 g/mL)
≈ 48.82 mL
Ta-dah! The answer is approximately 48.82 mL.
Remember, chemistry is all about precision, but let's not forget to have a laugh while we're at it!
To determine the volume of concentrated HCl solution needed to make 3.7 L of 0.435 M HCl, we can use the equation:
M1V1 = M2V2
Where:
M1 = initial concentration of HCl solution (37.0% by mass HCl)
V1 = volume of HCl solution needed
M2 = final concentration of HCl solution (0.435 M)
V2 = final volume of HCl solution (3.7 L)
First, we need to convert the final volume from liters to milliliters:
V2 = 3.7 L x 1000 mL/L
V2 = 3700 mL
Now, we can substitute the values into the equation and solve for V1:
37.0% (concentration) x V1 = 0.435 M (final concentration) x 3700 mL
0.37 V1 = 1609.5
V1 = 1609.5 / 0.37
V1 ≈ 4354.05 mL
Therefore, the correct amount of concentrated HCl solution needed to make 3.7 L of 0.435 M HCl is approximately 4354.05 mL.
To find the volume of concentrated HCl solution needed to make 3.7 L of 0.435 M HCl, we can use the equation:
M1V1 = M2V2
Where:
M1 = initial concentration of the concentrated HCl solution
V1 = volume of the concentrated HCl solution needed
M2 = final concentration of HCl solution (0.435 M)
V2 = final volume of the HCl solution (3.7 L)
First, let's convert the final volume to milliliters (mL):
3.7 L * 1000 mL/L = 3700 mL
Now, let's plug in the values into the equation and solve for V1 (volume of concentrated HCl solution needed):
37.0% by mass HCl = 37.0 g HCl / 100 g solution
Given that the density of the solution is 1.20 g/mL, we can calculate the volume of the concentrated HCl solution as follows:
37.0 g HCl / 100 g solution * V1 = 0.435 M * 3700 mL
To find V1, rearrange the equation:
V1 = (0.435 M * 3700 mL) / (37.0 g HCl / 100 g solution)
V1 = (0.435 M * 3700 mL) / (0.37)
V1 = (1609.5) / (0.37)
V1 ≈ 4351 mL
Therefore, the correct amount of the concentrated HCl solution needed to make 3.7 L of 0.435 M HCl is approximately 4351 mL.