To solve this problem, we use the concept of dilution and the formula for concentration:

$C_1 V_1 = C_2 V_2$

Where:

• $C_1$ is the initial concentration (0.5 M),
• $V_1$ is the initial volume (1 liter),
• $C_2$ is the final concentration (what we are solving for),
• $V_2$ is the final volume (2 liters).

Now, rearrange the formula to solve for $C_2$:

$C_2 = \frac{C_1 V_1}{V_2}$

Substitute the values:

$C_2 = \frac{0.5 \, \text{M} \times 1 \, \text{L}}{2 \, \text{L}} = \frac{0.5}{2} = 0.25 \, \text{M}$

So, the concentration of the final solution will be 0.25 M.

Would you like more details on the solution process?

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Here are some related questions to further your understanding:

1. How would the concentration change if 3 liters were the final volume instead of 2 liters?
2. What would happen if the initial solution were 1 M instead of 0.5 M?
3. How can you calculate the mass of solute in the final solution if you know the molar mass?
4. What if a certain amount of solute is added instead of dilution by volume—how does this affect concentration?
5. How does temperature affect the molarity of a solution?

Tip: When diluting a solution, remember that the amount of solute remains constant—only the volume changes.