Question
How many ml of 1% solution do you need, if you need to prepare 0.75% solution using 1% and 0.5% stock solutions?
Use the alligation method to solve this problem.
Use the alligation method to solve this problem.
Answers
GPT-4o mini
To solve the problem using the alligation method, we first need to set up our concentrations with respect to the desired concentration and the available stock solutions.
1. **Identify the Concentrations**:
- Let \( A \) be the 1% solution.
- Let \( B \) be the 0.5% solution.
- We want to prepare \( C \), a 0.75% solution.
2. **Set Up the Alligation**:
- The alligation method involves placing the concentrations in a diagram:
\[
\begin{array}{c|c}
\text{Concentration} & \text{Difference} \\
\hline
A (1\%) & C (0.75\%) - A (1\%) \to 0.75 - 1 = -0.25 \quad (\text{absolute value}) \\
C (0.75\%) & \\
B (0.5\%) & C (0.75\%) - B (0.5\%) \to 0.75 - 0.5 = 0.25 \\
\end{array}
\]
Now, we compute the differences:
- The absolute difference from \( A \) to \( C \) is \( |0.75 - 1| = 0.25 \).
- The absolute difference from \( B \) to \( C \) is \( |0.75 - 0.5| = 0.25 \).
3. **Ratio of Parts**:
- Place the absolute differences:
\[
\begin{array}{c|c}
\text{Solution} & \text{Difference} \\
\hline
A (1\%) & 0.25 \\
C (0.75\%) & \\
B (0.5\%) & 0.25 \\
\end{array}
\]
This gives us a 1:1 ratio of parts.
4. **Calculating Amounts**:
Let \( x \) be the volume of the 1% solution (A) and \( y \) be the volume of the 0.5% solution (B).
From the ratio, we know:
\[
\frac{x}{y} = 1 \quad \Rightarrow \quad x = y
\]
Let \( x + y = V \) be the total volume needed. Since \( x = y \), we can substitute:
\[
x + x = V \quad \Rightarrow \quad 2x = V \quad \Rightarrow \quad x = \frac{V}{2} \quad \text{and} \quad y = \frac{V}{2}
\]
5. **Final Volume**:
Choose a total volume \( V \). For instance, if we want a total volume of \( V = 100 \) ml as an example:
- Then, \( x = y = \frac{100}{2} = 50 \) ml.
Thus, to prepare a 0.75% solution from a 1% and a 0.5% solution, you would need:
- **50 ml of 1% solution**,
- **50 ml of 0.5% solution**.
In summary, for any given total volume \( V \), you would use \( \frac{V}{2} \) ml of each stock solution.
1. **Identify the Concentrations**:
- Let \( A \) be the 1% solution.
- Let \( B \) be the 0.5% solution.
- We want to prepare \( C \), a 0.75% solution.
2. **Set Up the Alligation**:
- The alligation method involves placing the concentrations in a diagram:
\[
\begin{array}{c|c}
\text{Concentration} & \text{Difference} \\
\hline
A (1\%) & C (0.75\%) - A (1\%) \to 0.75 - 1 = -0.25 \quad (\text{absolute value}) \\
C (0.75\%) & \\
B (0.5\%) & C (0.75\%) - B (0.5\%) \to 0.75 - 0.5 = 0.25 \\
\end{array}
\]
Now, we compute the differences:
- The absolute difference from \( A \) to \( C \) is \( |0.75 - 1| = 0.25 \).
- The absolute difference from \( B \) to \( C \) is \( |0.75 - 0.5| = 0.25 \).
3. **Ratio of Parts**:
- Place the absolute differences:
\[
\begin{array}{c|c}
\text{Solution} & \text{Difference} \\
\hline
A (1\%) & 0.25 \\
C (0.75\%) & \\
B (0.5\%) & 0.25 \\
\end{array}
\]
This gives us a 1:1 ratio of parts.
4. **Calculating Amounts**:
Let \( x \) be the volume of the 1% solution (A) and \( y \) be the volume of the 0.5% solution (B).
From the ratio, we know:
\[
\frac{x}{y} = 1 \quad \Rightarrow \quad x = y
\]
Let \( x + y = V \) be the total volume needed. Since \( x = y \), we can substitute:
\[
x + x = V \quad \Rightarrow \quad 2x = V \quad \Rightarrow \quad x = \frac{V}{2} \quad \text{and} \quad y = \frac{V}{2}
\]
5. **Final Volume**:
Choose a total volume \( V \). For instance, if we want a total volume of \( V = 100 \) ml as an example:
- Then, \( x = y = \frac{100}{2} = 50 \) ml.
Thus, to prepare a 0.75% solution from a 1% and a 0.5% solution, you would need:
- **50 ml of 1% solution**,
- **50 ml of 0.5% solution**.
In summary, for any given total volume \( V \), you would use \( \frac{V}{2} \) ml of each stock solution.