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This is a dilution problem, and we can solve it using the dilution equation:

\[
M_1 V_1 = M_2 V_2
\]

Where:
- \( M_1 \) is the initial concentration (2.10 M),
- \( V_1 \) is the initial volume (59.0 mL or 0.0590 L),
- \( M_2 \) is the final concentration (1.00 M),
- \( V_2 \) is the final volume (what we are solving for).

Now, let's rearrange the equation to solve for \( V_2 \):

\[
V_2 = \frac{M_1 V_1}{M_2}
\]

Substitute the known values:

\[
V_2 = \frac{(2.10 \, \text{M}) (0.0590 \, \text{L})}{1.00 \, \text{M}} = 0.1239 \, \text{L}
\]

Thus, the final volume \( V_2 \) is 0.124 L when rounded to three significant digits.

Would you like more details or have any further questions?

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Here are five related questions:
1. How does changing the concentration affect the final volume in dilution?
2. What happens to the moles of solute during dilution?
3. How do we know how much water to add during dilution?
4. What would happen if the initial concentration was higher?
5. Why is it important to maintain significant figures in scientific calculations?

**Tip:** Always remember to convert all units to liters (L) when working with volumes in molarity calculations.

A chemist must dilute 59.0 mL of 2.10 M aqueous sodium carbonate (Na₂CO₃) solution until the concentration falls to 1.00 M. She'll do this by adding distilled water to the solution until it reaches a certain final volume. Calculate this final volume, in liters.

Learn how to calculate the final volume required to dilute a 59.0 mL sample of 2.10 M aqueous sodium carbonate (Na₂CO₃) to 1.00 M. This problem covers key concepts like the dilution equation and molarity calculations, suitable for high school chemistry students.

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