What volumes of 95% ethanol and deionized water would you use to make 0.5 liters of a 19%
ethanol solution?

To prepare 0.5 liters of a 19% ethanol solution using 95% ethanol and deionized water, we can use the dilution formula:

\[
C_1 V_1 = C_2 V_2
\]

Where:
- \( C_1 \) is the concentration of the stock solution (95% ethanol),
- \( V_1 \) is the volume of the stock solution we need,
- \( C_2 \) is the concentration of the desired solution (19% ethanol),
- \( V_2 \) is the final volume of the desired solution (0.5 liters or 500 mL).

### Step-by-Step Calculation:

1. Plug the values into the dilution formula:

\[
95\% \times V_1 = 19\% \times 500 \, \text{mL}
\]

2. Solve for \( V_1 \):

\[
V_1 = \frac{19 \times 500}{95} = \frac{9500}{95} = 100 \, \text{mL}
\]

So, you need 100 mL of 95% ethanol.

3. The remaining volume (to reach 500 mL) will be deionized water:

\[
V_{\text{water}} = 500 \, \text{mL} - 100 \, \text{mL} = 400 \, \text{mL}
\]

### Final answer:
- You will need **100 mL of 95% ethanol** and **400 mL of deionized water** to make 0.5 liters of a 19% ethanol solution.

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Would you like more details or have any questions about this process?

Here are 5 related questions you can explore:
1. How would the calculation change for a 70% ethanol solution?
2. Can you calculate the mass of ethanol in the final solution?
3. How would the volume of water change for different final volumes?
4. What role does deionized water play in ethanol solutions?
5. How do you calculate concentration in weight/volume percentages?

**Tip**: When preparing solutions, always add the solute (ethanol) to the solvent (water) to avoid splashing and ensure even mixing.

What volumes of 95% ethanol and deionized water would you use to make 0.5 liters of a 19% ethanol solution?

Learn how to calculate the volumes of 95% ethanol and deionized water needed to prepare 0.5 liters of a 19% ethanol solution using the dilution formula. Step-by-step guide suitable for high school students, covering the concept of concentration and dilution.

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