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
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To answer these questions, we need to understand the concept of molarity, which measures the concentration of a solute in a solution.
a) What is the concentration of vinegar in mol/L?
To calculate the concentration of vinegar in mol/L, we need to convert the percentage of acetic acid to grams of acetic acid and then use the molar mass to convert it to moles.
Here's how to do it step-by-step:
1. Determine the mass of acetic acid in 100 g (assuming a 100 g sample of vinegar). Since the acetic acid concentration is given as 5.00% m/v, it means there are 5.00 grams of acetic acid in every 100 grams of vinegar.
2. Convert grams of acetic acid to moles. The molar mass of acetic acid (CH3COOH) is 60.05 g/mol. Divide the mass of acetic acid by its molar mass to get the number of moles.
moles = mass / molar mass = 5.00 g / 60.05 g/mol ≈ 0.083 mol
3. The volume of the vinegar solution is not given, so the concentration of vinegar in mol/L will be the same as the concentration of acetic acid in the concentrated solution. Therefore, the concentration of vinegar is approximately 17.5 mol/L.
b) What volume of vinegar can the cafeteria produce from a 4.00 L jug of concentrated acetic acid?
The volume of vinegar that can be produced depends on the dilution factor, which is the ratio of the initial concentration to the final concentration.
Here's how to calculate the volume of vinegar produced:
1. Determine the dilution factor. The dilution factor is given by the equation:
C1V1 = C2V2
Where C1 is the initial concentration, V1 is the initial volume, C2 is the final concentration, and V2 is the final volume.
In this case, the initial concentration (C1) is 17.5 mol/L, the initial volume (V1) is 4.00 L, the final concentration (C2) is not given, and the final volume (V2) is what we need to calculate.
Rearrange the equation to solve for V2:
V2 = (C1V1) / C2
2. Substitute the values into the equation:
V2 = (17.5 mol/L) * (4.00 L) / C2
3. We don't have the value of C2 given. However, since the concentration of vinegar is the same as the concentration of acetic acid, we can use the initial concentration of acetic acid (17.5 mol/L) as C2.
V2 = (17.5 mol/L) * (4.00 L) / 17.5 mol/L ≈ 4.00 L
Therefore, the cafeteria can produce approximately 4.00 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 answer this question, we need to determine the volume of water required to achieve the desired concentration. We can use the dilution equation again:
Here's how to calculate the volume of water needed:
1. Determine the dilution factor. The dilution factor is given by the equation:
C1V1 = C2V2
Where C1 is the initial concentration, V1 is the initial volume, C2 is the final concentration, and V2 is the final volume.
In this case, the initial concentration (C1) is 17.5 mol/L, the final concentration (C2) is 5.00% m/v, the initial volume (V1) is not given, and the final volume (V2) is the sum of the initial volume and the volume of water added (V1 + V_water).
The final concentration needs to be converted to mol/L before substitution:
Convert the percentage to grams:
mass of acetic acid in 100 g of vinegar = (5.00 g/100 g) * 100 g ≈ 5.00 g
Set up the equation to convert grams to moles:
moles = mass / molar mass = 5.00 g / 60.05 g/mol ≈ 0.083 mol
Now convert the moles to mol/L:
C2 = 0.083 mol / (V1 + V_water) L
2. Substitute the values into the equation:
(17.5 mol/L) * V1 = (0.083 mol) / (V1 + V_water)
3. Solve for V_water:
V_water = (0.083 mol / 17.5 mol/L) - V1
4. We don't have the value of V1 given. However, since the total volume of the solution is the sum of the volume of acetic acid and the volume of water, we can assume that V1 = 4.00 L (volume of acetic acid from a 4.00 L jug).
V_water = (0.083 mol / 17.5 mol/L) - 4.00 L
Therefore, the cafeteria would need to add approximately (0.083 mol / 17.5 mol/L) - 4.00 L of water to the concentrated acetic acid.