I have no idea where to even start with this problem on my pre-lab!

For part of the experiment, you will need to prepare 100 mL of 0.025 M CH3CO2H, starting from commercial "glacial" acetic acid, which has a concentration of 17.4 M. This is best done by making two dilutions, in order to avoid either having to measure out an extremely small volume of glacial acetic acid, or having to make a very large volume of diluted acid. For example, you could measure out 1.00 mL of glacial acetic acid, dilute it to 50.00 mL in a 50 mL volumetric flask, and then take 7.2 mL of this solution, and it dilute it to 100.00 mL in a 100 mL volumetric flask.

Verify this procedure will result in 100 mL of 0.025 M acetic acid.

Part 2:
Devise a method similar to that in the previous question to make 100.00 mL of 0.025 M NH3, from concentrated ammonia, which has a concentration of 14.5 M?

Part 3:
Consider the conductance of the following 0.025 M solutions. Use the data to classify each solute as either a nonelectrolyte, a weak electrolyte, or a strong electrolyte:
Solute, conductance (mS), SE/WE/NE
A, 3.5, ?
B, 7.6, ?
C, 0.0, ?
D, 0.2, ?
E, 6.5, ?
F, 0.3, ?
G, 2.9, ?

a. 17.4 M x (1.00mL/50.00mL) = 0.348M

Then 0.348 x (7.2 mL/100 mL ) = 0.025M

b. Use a as a template.
c. I would call the 7.6 and 6.5 SE, the zeros are NE (of course) while 3.5 and 2.9 are WE. I don't know how you want to handle athe 0.2 and 0.3. Those could be non-electrolytes that had a small contamination OR they could be very weak solution of WE.

Thanks so much

To prepare 100 mL of 0.025 M CH3CO2H (acetic acid) from commercial "glacial" acetic acid with a concentration of 17.4 M, you can follow these steps:

Step 1: Calculate the volume needed for the first dilution.
Use the concept of dilution formula:
C1V1 = C2V2
Where:
C1 = Initial concentration
V1 = Volume used
C2 = Final concentration
V2 = Final volume

Let's calculate the volume needed for the first dilution:
C1 = 17.4 M
V1 = ?
C2 = 0.025 M
V2 = 50 mL (since it will be diluted in a 50 mL volumetric flask)

(17.4 M)(V1) = (0.025 M)(50 mL)
V1 = (0.025 M)(50 mL) / (17.4 M)
V1 ≈ 0.072 mL (rounded to three decimal places)

So, you need approximately 0.072 mL for the first dilution.

Step 2: Perform the first dilution.
Measure out 0.072 mL of glacial acetic acid and add it to a 50 mL volumetric flask.
Then, add distilled water to the 50 mL mark and mix well. This will give you a diluted solution.

Step 3: Calculate the volume needed for the second dilution.
Use the same dilution formula:
C1V1 = C2V2
Where:
C1 = Concentration after the first dilution (0.072 mL)
V1 = Volume used after the first dilution (50 mL)
C2 = Final concentration (0.025 M)
V2 = Final volume (100 mL)

(0.072 M)(50 mL) = (0.025 M)(V2)
V2 = (0.072 M)(50 mL) / (0.025 M)
V2 ≈ 144 mL (rounded to three decimal places)

So, you need approximately 144 mL for the second dilution.

Step 4: Perform the second dilution.
Measure out 7.2 mL of the diluted solution from the first dilution and add it to a 100 mL volumetric flask.
Then, add distilled water to the 100 mL mark and mix well. This will give you a final volume of 100 mL of 0.025 M CH3CO2H (acetic acid).

Now, let's move on to Part 2:

To make 100.00 mL of 0.025 M NH3 (ammonia) from concentrated ammonia with a concentration of 14.5 M, you can follow a similar procedure as in Part 1:

Step 1: Calculate the volume needed for the first dilution.
Use the dilution formula:
C1V1 = C2V2

Let's calculate the volume needed for the first dilution:
C1 = 14.5 M
V1 = ?
C2 = 0.025 M
V2 = 50 mL (since it will be diluted in a 50 mL volumetric flask)

(14.5 M)(V1) = (0.025 M)(50 mL)
V1 = (0.025 M)(50 mL) / (14.5 M)
V1 ≈ 0.086 mL (rounded to three decimal places)

So, you need approximately 0.086 mL for the first dilution.

Step 2: Perform the first dilution.
Measure out 0.086 mL of concentrated ammonia and add it to a 50 mL volumetric flask.
Then, add distilled water to the 50 mL mark and mix well. This will give you a diluted solution.

Step 3: Calculate the volume needed for the second dilution.
Use the same dilution formula:
C1V1 = C2V2

(0.086 M)(50 mL) = (0.025 M)(V2)
V2 = (0.086 M)(50 mL) / (0.025 M)
V2 ≈ 172 mL (rounded to three decimal places)

So, you need approximately 172 mL for the second dilution.

Step 4: Perform the second dilution.
Measure out 17.2 mL of the diluted solution from the first dilution and add it to a 100 mL volumetric flask.
Then, add distilled water to the 100 mL mark and mix well. This will give you a final volume of 100 mL of 0.025 M NH3 (ammonia).

Moving on to Part 3:

To classify each solute as a nonelectrolyte, weak electrolyte, or strong electrolyte based on their conductance, use the following guidelines:

- Nonelectrolyte: Substances that do not ionize in solution, therefore have very low conductance (close to 0 mS).
- Weak electrolyte: Substances that partially ionize in solution, therefore have moderate conductance.
- Strong electrolyte: Substances that completely ionize in solution, therefore have high conductance.

Based on the provided conductance data:

A, 3.5 mS: Weak electrolyte
B, 7.6 mS: Weak electrolyte
C, 0.0 mS: Nonelectrolyte
D, 0.2 mS: Weak electrolyte
E, 6.5 mS: Weak electrolyte
F, 0.3 mS: Weak electrolyte
G, 2.9 mS: Weak electrolyte

Therefore, the classifications are as follows:
A: Weak electrolyte
B: Weak electrolyte
C: Nonelectrolyte
D: Weak electrolyte
E: Weak electrolyte
F: Weak electrolyte
G: Weak electrolyte

To solve this problem, let's break it down into each part.

Part 1:
To prepare 100 mL of 0.025 M CH3CO2H from glacial acetic acid (17.4 M), we can use the dilution method described. Here's how you can verify if this procedure will result in 100 mL of 0.025 M acetic acid:

1. Start by calculating the moles of CH3CO2H needed:
Moles = Molarity × Volume (in liters)
Moles = 0.025 M × 0.100 L

2. Calculate the volume of glacial acetic acid needed:
Volume of glacial acetic acid = Moles of CH3CO2H / Concentration of glacial acetic acid
Volume of glacial acetic acid = (0.025 M × 0.100 L) / 17.4 M

3. Follow the dilution steps mentioned in the question:
a. Measure out 1.00 mL of glacial acetic acid.
b. Dilute it to 50.00 mL in a 50 mL volumetric flask.
c. Take 7.2 mL of this solution and dilute it to 100.00 mL in a 100 mL volumetric flask.

4. Now, calculate the final concentration of the acetic acid solution:
Concentration = Moles / Volume (in liters)
Concentration = (0.025 M × 0.100 L) / 0.100 L

Compare this value to the desired concentration of 0.025 M. If they match, the procedure is correct.

Part 2:
Similarly, let's devise a method to make 100.00 mL of 0.025 M NH3 from concentrated ammonia (14.5 M):

1. Calculate the moles of NH3 needed:
Moles = Molarity × Volume (in liters)
Moles = 0.025 M × 0.100 L

2. Calculate the volume of concentrated ammonia needed:
Volume of concentrated ammonia = Moles of NH3 / Concentration of ammonia
Volume of concentrated ammonia = (0.025 M × 0.100 L) / 14.5 M

3. Follow the same dilution method as described above using volumetric flasks.

Part 3:
To classify each solute as either a nonelectrolyte, weak electrolyte, or strong electrolyte based on the provided conductance data:

- A nonelectrolyte will typically have a low or zero conductance (conductivity) because it does not dissociate into ions in solution.
- A weak electrolyte will have moderate conductance due to partial dissociation into ions.
- A strong electrolyte will have high conductance because it completely dissociates into ions in solution.

By comparing the conductance values with this understanding:
- A: 3.5 mS -> ?
- B: 7.6 mS -> ?
- C: 0.0 mS -> ?
- D: 0.2 mS -> ?
- E: 6.5 mS -> ?
- F: 0.3 mS -> ?
- G: 2.9 mS -> ?

Based on the provided data, you can determine the classification of each solute as SE (strong electrolyte), WE (weak electrolyte), or NE (nonelectrolyte).