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.