A chemist has on hand, a supply of a 50% methyl alcohol solution and 16% methyl alcohol solution. They want to create a final solution of 510 mL with a concentration of 30%.

b. How much of each solution should be mixed together to create the final solution.

To solve this problem, we need to determine the amount of each solution (50% methyl alcohol and 16% methyl alcohol) that should be mixed to obtain 510 mL of a 30% methyl alcohol solution. We can set up a system of two equations to find the required amounts.

### Define the Variables:
- Let \( x \) be the volume (in mL) of the 50% methyl alcohol solution.
- Let \( y \) be the volume (in mL) of the 16% methyl alcohol solution.

### Set Up the Equations:
1. **Total Volume Equation:**
   The total volume of the mixture should be 510 mL. So, we have:
   \[
   x + y = 510 \quad \text{(1)}
   \]

2. **Total Methyl Alcohol Equation:**
   The total amount of methyl alcohol in the mixture is the sum of the methyl alcohol from both solutions, which should equal 30% of 510 mL. This can be expressed as:
   \[
   0.50x + 0.16y = 0.30 \times 510 \quad \text{(2)}
   \]

### Simplifying the Equations:
First, calculate the right-hand side of equation (2):
\[
0.30 \times 510 = 153 \quad \text{(mL of pure methyl alcohol in the final solution)}
\]
So, the system of equations becomes:

\[
\begin{aligned}
x + y &= 510 \quad \text{(1)} \\
0.50x + 0.16y &= 153 \quad \text{(2)}
\end{aligned}
\]

### Solve the System:
Now, let's solve this system of equations.

**Step 1: Solve Equation (1) for \( y \):**
\[
y = 510 - x \quad \text{(3)}
\]

**Step 2: Substitute Equation (3) into Equation (2):**
\[
0.50x + 0.16(510 - x) = 153
\]

**Step 3: Distribute and simplify:**
\[
0.50x + 81.6 - 0.16x = 153
\]
\[
0.34x + 81.6 = 153
\]

**Step 4: Solve for \( x \):**
\[
0.34x = 153 - 81.6
\]
\[
0.34x = 71.4
\]
\[
x = \frac{71.4}{0.34} \approx 210 \, \text{mL}
\]

**Step 5: Substitute \( x = 210 \) mL back into Equation (3) to find \( y \):**
\[
y = 510 - 210 = 300 \, \text{mL}
\]

### Final Answer:
- \( x = 210 \) mL of the 50% methyl alcohol solution.
- \( y = 300 \) mL of the 16% methyl alcohol solution.

So, the chemist should mix 210 mL of the 50% solution with 300 mL of the 16% solution to obtain 510 mL of a 30% methyl alcohol solution.

Do you have any further questions, or would you like to see more details?

Here are 8 related questions to deepen your understanding:

1. How can you verify that the final solution indeed has a 30% concentration?
2. What if the final solution needed to be 40% methyl alcohol? How would the volumes change?
3. How would you solve the system using the elimination method?
4. How does the total volume affect the concentrations in the final mixture?
5. What are the implications if the available concentrations were different, say 40% and 20%?
6. How would the problem change if the chemist had only 400 mL of one of the solutions available?
7. How do you interpret the coefficients in the system of equations in the context of this problem?
8. What are other real-world scenarios where a similar mixture problem might arise?

**Tip:** When working with percentages in mixture problems, remember that percentages must be converted to decimals (e.g., 50% = 0.50) when used in equations.

Learn how to create a 30% methyl alcohol solution by mixing 50% and 16% methyl alcohol solutions. Follow step-by-step calculations to determine the volumes needed for each solution. Suitable for high school students studying mixture problems in chemistry.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation