Answered

You are given a 0.05 mol/L solution of CrCl₃. How would you use this solution to prepare 100 mL quantities of solutions with concentrations of 0.04, 0.03, 0.02, and 0.01 mol/L?



Answer :

To prepare 100 mL quantities of solutions with concentrations of 0.04 mol/L, 0.03 mol/L, 0.02 mol/L, and 0.01 mol/L from a 0.05 mol/L CrCl3 solution, we will use the dilution formula:

[tex]\[ C_1 \times V_1 = C_2 \times V_2 \][/tex]

where:
- [tex]\( C_1 \)[/tex] is the initial concentration (0.05 mol/L),
- [tex]\( V_1 \)[/tex] is the volume of the initial solution needed,
- [tex]\( C_2 \)[/tex] is the desired concentration,
- [tex]\( V_2 \)[/tex] is the final volume of the desired solution (0.1 L or 100 mL).

We will calculate [tex]\( V_1 \)[/tex] for each desired concentration and then determine the volume of water to be added to achieve the final volume of 100 mL.

### 1. Desired concentration: 0.04 mol/L

[tex]\[ C_1 \times V_1 = C_2 \times V_2 \][/tex]
[tex]\[ 0.05 \, \text{mol/L} \times V_1 = 0.04 \, \text{mol/L} \times 0.1 \, \text{L} \][/tex]
[tex]\[ V_1 = \frac{0.04 \times 0.1}{0.05} \][/tex]
[tex]\[ V_1 = 0.08 \, \text{L} \, (80 \, \text{mL}) \][/tex]

Volume of water to add: [tex]\( 100 \, \text{mL} - 80 \, \text{mL} = 20 \, \text{mL} \)[/tex]

### 2. Desired concentration: 0.03 mol/L

[tex]\[ 0.05 \, \text{mol/L} \times V_1 = 0.03 \, \text{mol/L} \times 0.1 \, \text{L} \][/tex]
[tex]\[ V_1 = \frac{0.03 \times 0.1}{0.05} \][/tex]
[tex]\[ V_1 = 0.06 \, \text{L} \, (60 \, \text{mL}) \][/tex]

Volume of water to add: [tex]\( 100 \, \text{mL} - 60 \, \text{mL} = 40 \, \text{mL} \)[/tex]

### 3. Desired concentration: 0.02 mol/L

[tex]\[ 0.05 \, \text{mol/L} \times V_1 = 0.02 \, \text{mol/L} \times 0.1 \, \text{L} \][/tex]
[tex]\[ V_1 = \frac{0.02 \times 0.1}{0.05} \][/tex]
[tex]\[ V_1 = 0.04 \, \text{L} \, (40 \, \text{mL}) \][/tex]

Volume of water to add: [tex]\( 100 \, \text{mL} - 40 \, \text{mL} = 60 \, \text{mL} \)[/tex]

### 4. Desired concentration: 0.01 mol/L

[tex]\[ 0.05 \, \text{mol/L} \times V_1 = 0.01 \, \text{mol/L} \times 0.1 \, \text{L} \][/tex]
[tex]\[ V_1 = \frac{0.01 \times 0.1}{0.05} \][/tex]
[tex]\[ V_1 = 0.02 \, \text{L} \, (20 \, \text{mL}) \][/tex]

Volume of water to add: [tex]\( 100 \, \text{mL} - 20 \, \text{mL} = 80 \, \text{mL} \)[/tex]

### Summary
- To make 0.04 mol/L solution: Mix 80 mL of the initial solution with 20 mL of water.
- To make 0.03 mol/L solution: Mix 60 mL of the initial solution with 40 mL of water.
- To make 0.02 mol/L solution: Mix 40 mL of the initial solution with 60 mL of water.
- To make 0.01 mol/L solution: Mix 20 mL of the initial solution with 80 mL of water.