Describe how to make 750.0 mL of 0.1667 M NH4C2H3O2 from 0.225 M NH4C2H3O2?
dilution formula
and with this one I did
0.1667(750.0)=0.225(v)
125.025/0.225=555.7??
That's what I obtained. So you take 555.7 (that may be too many significant figures) of the 0.225M stuff and add H2O to 750 mL.
To make 750.0 mL of 0.1667 M NH4C2H3O2 from 0.225 M NH4C2H3O2, you will need to dilute the 0.225 M solution by adding a calculated amount of solvent (most commonly water) to achieve the desired concentration.
To calculate the required volume of the 0.225 M NH4C2H3O2 solution to be used, we can use the formula:
C1V1 = C2V2
Where:
C1 = initial concentration (0.225 M)
V1 = initial volume (unknown)
C2 = final concentration (0.1667 M)
V2 = final volume (750.0 mL or 0.750 L)
Rearranging the formula and substituting the values, we get:
V1 = (C2 * V2) / C1
V1 = (0.1667 M * 0.750 L) / 0.225 M
V1 ≈ 0.555 L or 555 mL
So, you need approximately 555 mL of the 0.225 M NH4C2H3O2 solution.
To prepare the desired 0.1667 M NH4C2H3O2 solution, follow these steps:
1. Measure 555 mL of the 0.225 M NH4C2H3O2 solution using a graduated cylinder or a volumetric flask.
2. Pour the measured 0.225 M NH4C2H3O2 solution into a container.
3. Add water to the container to bring the total volume up to 750.0 mL (0.750 L). You can use a graduated cylinder or a volumetric flask to accurately measure the required volume of water.
4. Mix the solution thoroughly to ensure complete homogeneity.
Now you have successfully prepared 750.0 mL of 0.1667 M NH4C2H3O2 solution.