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What volume (in [tex]L[/tex]) of a 1.25 M potassium fluoride (KF) solution would be needed to make 455 mL of a 0.838 M solution by dilution?

Sagot :

Sure, let me guide you step-by-step through the process of solving this problem:

### Given Data:
1. Initial molarity ([tex]\(M_1\)[/tex]) of the potassium fluoride (KF) solution: 1.25 M
2. Final volume ([tex]\(V_2\)[/tex]) of the desired solution: 455 mL
3. Desired molarity ([tex]\(M_2\)[/tex]) of the final solution: 0.838 M

Our goal is to find the volume ([tex]\(V_1\)[/tex]) of the initial solution needed to achieve the desired concentration and volume through dilution.

### Step 1: Understand the formula for dilution
We use the dilution equation for this purpose:
[tex]\[ M_1 \times V_1 = M_2 \times V_2 \][/tex]

### Step 2: Substitute the given values
We substitute the known values into the equation to find [tex]\(V_1\)[/tex]:
[tex]\[ 1.25 \times V_1 = 0.838 \times 455 \][/tex]

### Step 3: Solve for [tex]\(V_1\)[/tex]
First, calculate the right-hand side:
[tex]\[ 0.838 \times 455 = 381.29 \][/tex]

Next, solve for [tex]\(V_1\)[/tex]:
[tex]\[ V_1 = \frac{381.29}{1.25} \][/tex]

[tex]\[ V_1 \approx 305.032 \, \text{mL} \][/tex]

### Step 4: Convert the volume from mL to L
To convert from milliliters to liters, we use the conversion factor [tex]\(1 \, \text{L} = 1000 \, \text{mL}\)[/tex]:
[tex]\[ V_1 = \frac{305.032 \, \text{mL}}{1000} \][/tex]

[tex]\[ V_1 \approx 0.305 \, \text{L} \][/tex]

### Conclusion:
The volume of the 1.25 M potassium fluoride (KF) solution needed to make 455 mL of a 0.838 M solution is approximately 305.032 mL, or equivalently, 0.305 L.