A student delivered 19.01 mL (V1) of 3.19% stock solution (M1) using a graduated pipette into a 50 mL (V2) flask. Distilled water was added to the flask to the graduation mark and mixed well. Using the dilution equation, calculate the concentration of the diluted solution.

To calculate the concentration of the diluted solution, we can use the dilution equation:

\[
M_1 V_1 = M_2 V_2
\]

Where:
- \( M_1 \) is the initial concentration (stock solution) = 3.19%
- \( V_1 \) is the volume of the stock solution = 19.01 mL
- \( M_2 \) is the concentration of the diluted solution (what we want to find)
- \( V_2 \) is the final total volume after dilution = 50 mL

We can rearrange the equation to solve for \( M_2 \):

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

Now substitute the known values:

\[
M_2 = \frac{(3.19\%) \times (19.01 \, \text{mL})}{50 \, \text{mL}}
\]

Let me calculate this.The concentration of the diluted solution, \( M_2 \), is approximately 1.21%.

Would you like more details on this calculation or have any questions?

Here are 5 related questions to expand your understanding:
1. How would the concentration change if the final volume was 100 mL instead of 50 mL?
2. What would happen to the concentration if the initial volume (\( V_1 \)) was reduced to 10 mL?
3. How do you convert percentages into molar concentration in solutions?
4. How would this equation change if you were diluting a solution of molarity instead of percentage?
5. What is the practical use of dilution equations in chemical laboratories?

**Tip:** Always make sure to use consistent units when applying the dilution equation.

Learn how to calculate the concentration of a diluted solution using the dilution equation. This problem involves 19.01 mL of a 3.19% stock solution diluted to a final volume of 50 mL. Ideal for students in Grades 10-12.

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