The cafeteria decides to save money by using concentrated acetic acid (CH3COOH at 17.5 mol/L) and diluting it with water toproduce vinegar (5.00% m/v aceticacid).
a) What is the concentration of vinegar in mol/L?b) What volume of vinegar can the cafeteria produce from a 4.00 L jug of concentrated acetic acid?c) How much water would they need to add to the concentrated acet
a) To find the concentration of vinegar in mol/L, we need to convert the percentage of acetic acid to mol/L.
Given:
Concentration of concentrated acetic acid = 17.5 mol/L
Percentage of acetic acid in vinegar = 5.00% m/v
To calculate the concentration in mol/L, we need to know the density of the solution. Assuming the density of vinegar is approximately 1 g/mL, we can convert the percentage to mol/L.
First, we need to convert the percentage to grams of acetic acid in 100 mL (assuming 100 mL of vinegar):
5.00% m/v = 5.00 g/100 mL
Next, we can convert the grams of acetic acid to moles using the molar mass of acetic acid (60.05 g/mol):
5.00 g / 60.05 g/mol = 0.083 mol
Since we assumed 100 mL vinegar, we can calculate the concentration in mol/L:
0.083 mol / 0.1 L = 0.83 mol/L
Therefore, the concentration of vinegar is approximately 0.83 mol/L.
b) To find the volume of vinegar that can be produced from a 4.00 L jug of concentrated acetic acid, we need to use the concentration of the acid vinegar.
Given:
Concentration of vinegar = 0.83 mol/L
Volume of concentrated acetic acid = 4.00 L
Using the equation:
M1V1 = M2V2
M1 = concentration of concentrated acetic acid = 17.5 mol/L
V1 = volume of concentrated acetic acid (4.00 L)
M2 = concentration of vinegar = 0.83 mol/L
V2 = volume of vinegar (what we want to find)
Substituting the values into the equation:
17.5 mol/L × 4.00 L = 0.83 mol/L × V2
70 = 0.83V2
V2 = 70 / 0.83 ≈ 84.3 L
Therefore, the cafeteria can produce approximately 84.3 L of vinegar from a 4.00 L jug of concentrated acetic acid.
c) To find out how much water they need to add to the concentrated acetic acid, we need to calculate the volume difference between the concentrated acetic acid and the vinegar.
Given:
Volume of concentrated acetic acid = 4.00 L
Volume of vinegar = 84.3 L
The volume of water added = Volume of vinegar - Volume of concentrated acetic acid
Volume of water = 84.3 L - 4.00 L = 80.3 L
Therefore, the cafeteria would need to add approximately 80.3 L of water to the concentrated acetic acid.
To find the answers to these questions, we need to use the concept of concentration and dilution calculations.
a) What is the concentration of vinegar in mol/L?
First, let's convert the 5.00% m/v concentration of acetic acid in vinegar to mol/L.
Step 1: Convert the percentage to grams per liter (g/L)
In this case, 5.00% means 5.00 grams of acetic acid in 100 mL of vinegar. To convert to g/L, we need to multiply by 10 to get the concentration per liter.
5.00 g/100 mL * 10 = 5.00 g/L
Step 2: Convert grams to moles using the molar mass of acetic acid (CH3COOH)
The molar mass of acetic acid is as follows:
C = 12.01 g/mol
H = 1.01 g/mol (x3 for 3 hydrogens)
O = 16.00 g/mol (x2 for 2 oxygens)
Molar mass of acetic acid = (12.01 * 2) + (1.01 * 3) + 16.00 = 60.05 g/mol
Now, we can convert the grams of acetic acid to moles:
5.00 g / 60.05 g/mol = 0.083 moles
Therefore, the concentration of acetic acid in vinegar is 0.083 mol/L.
b) What volume of vinegar can the cafeteria produce from a 4.00 L jug of concentrated acetic acid?
For this, we need to use the concentration and volume relationship between the concentrated acetic acid and the vinegar.
The initial concentration of the concentrated acetic acid is given as 17.5 mol/L.
Since the concentration remains the same after dilution, we can set up the following equation:
Initial volume * initial concentration = Final volume * final concentration
(4.00 L) * (17.5 mol/L) = Final volume * (0.083 mol/L)
Rearranging the equation to solve for the final volume:
Final volume = (4.00 L * 17.5 mol/L) / 0.083 mol/L
Final volume = 840 L
Therefore, the cafeteria can produce 840 L of vinegar from a 4.00 L jug of concentrated acetic acid.
c) How much water would they need to add to the concentrated acetic acid?
To find the volume of water needed, we can subtract the volume of concentrated acetic acid from the final volume of vinegar.
Volume of water = Final volume - Initial volume
Volume of water = 840 L - 4.00 L
Volume of water = 836 L
Therefore, the cafeteria would need to add 836 L of water to the concentrated acetic acid to produce vinegar.
and, always add acid to water.
a. Lets change five percent to moles /liter. Assume in the dilute solution, the density is 1kg/liter
volume:1 liter, 1000g,
mass of vinegar: .05*1000=50g
moles/liter= 50/molmass*1=5/6=.73 check that.
So your are diluting it 17.5/.73 times, or 24times. That means one part conc acid, then 23 parts water.
So you are given 4L acid, so you need 23*4L water.