How do I prepare 80 mL of 0.10 M HC2H3O2 from a 6.0 M solution of Acetic acid?
How do I prepare 80 mL of 0.10 M NH3 from 6.0 M ammonia?
How accurately is the acetic acid solution to be made?
mL x M = mL x M
80 mL x 0.1M = mL x 6.0M
mL = (80*0.1/6.0) = 1.33
Technically, the way you do it is to dispense 1.33 mL of the stock solution into an 80 mL volumetric flask, then make the solution to the mark with distilled water. However, I don't know that I've seen an 80 mL volumetric flask.
Well, here's a fun chemistry joke for you: Why do chemists like nitrates so much? Because they're cheaper than day rates!
Now, let's get back to your question. To prepare 80 mL of a 0.10 M HC2H3O2 solution from a 6.0 M solution, we'll need to do some dilution. The formula for dilution is 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, C1 = 6.0 M, V1 is what we're trying to find, C2 = 0.10 M, and V2 = 80 mL. Plugging these values into the equation, we have:
(6.0 M)(V1) = (0.10 M)(80 mL)
Now, let's solve for V1:
V1 = (0.10 M)(80 mL) / 6.0 M
So, V1 = 1.33 mL. Therefore, to prepare 80 mL of a 0.10 M HC2H3O2 solution, you would measure out 1.33 mL of the 6.0 M acetic acid solution and then add enough solvent (like water) to reach a total volume of 80 mL.
Now, onto the second question about preparing 80 mL of 0.10 M NH3 solution from a 6.0 M ammonia solution. Let me think of a funny ammonia joke! Ah, here it is: Why was the ammonia molecule so cool? Because it had a chilling personality! Alright, now, let's solve this problem.
Using the same dilution formula, we can set up the equation:
(6.0 M)(V1) = (0.10 M)(80 mL)
Now, solving for V1:
V1 = (0.10 M)(80 mL) / 6.0 M
V1 is approximately equal to 1.33 mL. So, to prepare 80 mL of a 0.10 M NH3 solution, you would measure out 1.33 mL of the 6.0 M ammonia solution and then add enough solvent to reach a total volume of 80 mL.
I hope that helps! Let me know if you have any other questions or if you want to hear another joke.
To prepare 80 mL of 0.10 M HC2H3O2 (acetic acid) from a 6.0 M solution of acetic acid, you can follow these steps:
Step 1: Calculate the volume of the 6.0 M solution needed:
To do this, we can use the equation:
M1 x V1 = M2 x V2
Where:
M1 = initial concentration = 6.0 M
V1 = initial volume
M2 = final concentration = 0.10 M
V2 = final volume = 80 mL (or 0.080 L, since 1 mL = 0.001 L)
Substituting the values into the equation, we get:
6.0 M x V1 = 0.10 M x 0.080 L
Solving for V1:
V1 = (0.10 M x 0.080 L)/6.0 M
V1 ≈ 0.0013 L (or 1.3 mL)
Therefore, you would need to measure approximately 1.3 mL of the 6.0 M acetic acid solution.
Step 2: Add water to achieve the final volume:
Since you need to prepare 80 mL of the 0.10 M acetic acid solution, you will add water to bring the total volume to 80 mL. Measure out 78.7 mL (or 78.7 g) of water and add it to a container.
Step 3: Add the measured amount of the 6.0 M solution:
Next, add the 1.3 mL of the 6.0 M acetic acid solution to the container containing the water. Mix well to ensure the acetic acid is evenly distributed.
That's it! You have prepared 80 mL of 0.10 M HC2H3O2.
Now, let's move on to preparing 80 mL of 0.10 M NH3 (ammonia) from a 6.0 M ammonia solution:
Step 1: Calculate the volume of the 6.0 M solution needed:
Using the same formula as before (M1 x V1 = M2 x V2), we can calculate the volume of the 6.0 M solution needed.
M1 = 6.0 M (initial concentration)
V1 = ?
M2 = 0.10 M (final concentration)
V2 = 80 mL (or 0.080 L)
6.0 M x V1 = 0.10 M x 0.080 L
Solving for V1:
V1 = (0.10 M x 0.080 L)/6.0 M
V1 ≈ 0.0013 L (or 1.3 mL)
You would need to measure approximately 1.3 mL of the 6.0 M ammonia solution.
Step 2: Add water to achieve the final volume:
Again, add water to reach a final volume of 80 mL. Measure out 78.7 mL (or 78.7 g) of water and add it to a container.
Step 3: Add the measured amount of the 6.0 M solution:
Add the 1.3 mL of the 6.0 M ammonia solution to the container containing the water. Mix well to ensure the ammonia is evenly distributed.
Now, you have prepared 80 mL of 0.10 M NH3.
To prepare a solution with a specific molarity, you can use the formula:
M1 * V1 = M2 * V2
where M1 represents the initial molarity, V1 represents the initial volume, M2 represents the desired molarity, and V2 represents the desired volume.
For both the acetic acid solution and the ammonia solution, you need to dilute the 6.0 M solution to obtain the desired concentration.
For the acetic acid solution (HC2H3O2):
1. Identify the given molarity (M1) and volume (V1) of the 6.0 M solution. In this case, M1 = 6.0 M and V1 is not provided.
2. Identify the desired molarity (M2) and volume (V2) of the final solution. In this case, M2 = 0.10 M and V2 = 80 mL (or 0.080 L).
3. Rearrange the formula M1 * V1 = M2 * V2 to solve for V1: V1 = (M2 * V2) / M1.
4. Substitute the given values into the formula and calculate V1: V1 = (0.10 M * 0.080 L) / 6.0 M.
5. Calculate the volume V1 to obtain the initial volume required for dilution.
For the ammonia solution (NH3):
1. Follow the same steps as above, using the given values for the ammonia solution. In this case, M1 = 6.0 M, M2 = 0.10 M, and V2 = 80 mL (or 0.080 L).
2. Rearrange the formula M1 * V1 = M2 * V2 to solve for V1: V1 = (M2 * V2) / M1.
3. Substitute the given values into the formula and calculate V1: V1 = (0.10 M * 0.080 L) / 6.0 M.
By using these calculations, you can determine the initial volume required for dilution in order to prepare the desired concentration of both solutions.