Answer :

Answer:

To calculate the volume of 12M HCI needed to make the desired molarity solutions, use the formula:

M1V1 = M2V2

We know: M1 = 12M (molarity of stock solution) V1 = volume of stock solution needed M2 = desired molarity V2 = final volume (1.0L)

Let's calculate the volume of 12M HCI needed for each desired molarity solution:

For 0.50 M HCI solution: 12M * V1 = 0.5M * 1000mL V1 = (0.5M * 1000mL) / 12M V1 = 41.67mL

For 1.0 M HCl solution: 12M * V1 = 1.0M * 1000mL V1 = (1.0M * 1000mL) / 12M V2 = 83.33mL

For 1.5 M HCl solution: 12M * V1 = 1.5M * 1000mL V1 = (1.5M * 1000mL) / 12M V1 = 125mL

For 2.0 M HCl solution: 12M * V1 = 2.0M * 1000mL V1 = (2.0M * 1000mL) / 12M V1 = 166.67mL

For 5.0 M HCl solution: 12M * V1 = 5.0M * 1000mL V1 = (5.0M * 1000mL) / 12M V1 = 416.67mL

Using the dilution formula and the given concentrations, here are the calculated volumes:

1. **For 0.50M HCl solution:**

\[ V_1 = \frac{0.50 \, \text{M} \times 1.0 \, \text{L}}{12 \, \text{M}} = \frac{0.50}{12} = 0.0417 \, \text{L} = 41.7 \, \text{mL} \]

2. **For 1.0M HCl solution:**

\[ V_1 = \frac{1.0 \, \text{M} \times 1.0 \, \text{L}}{12 \, \text{M}} = \frac{1.0}{12} = 0.0833 \, \text{L} = 83.3 \, \text{mL} \]

3. **For 1.5M HCl solution:**

\[ V_1 = \frac{1.5 \, \text{M} \times 1.0 \, \text{L}}{12 \, \text{M}} = \frac{1.5}{12} = 0.125 \, \text{L} = 125 \, \text{mL} \]

4. **For 2.0M HCl solution:**

\[ V_1 = \frac{2.0 \, \text{M} \times 1.0 \, \text{L}}{12 \, \text{M}} = \frac{2.0}{12} = 0.167 \, \text{L} = 167 \, \text{mL} \]

5. **For 5.0M HCl solution:**

\[ V_1 = \frac{5.0 \, \text{M} \times 1.0 \, \text{L}}{12 \, \text{M}} = \frac{5.0}{12} = 0.417 \, \text{L} = 417 \, \text{mL} \]

So, the volumes needed are:

- **0.50M:** 41.7 mL

- **1.0M:** 83.3 mL

- **1.5M:** 125 mL

- **2.0M:** 167 mL

- **5.0M:** 417 mL

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