If a mixture is 6.25mg of M and 10mg of C per 5ml, how many ml of the mixture would 600mg of part C make?

To solve this, let's break the problem down.

We know:
- In 5 mL of the mixture, there is 10 mg of C (part C).
- We need to determine how many mL of the mixture would contain 600 mg of C.

### Step-by-Step Solution:

1. **Set up the ratio**:
   - 5 mL of the mixture contains 10 mg of C.
   - We want to know how many mL, \( x \), would contain 600 mg of C.

2. **Write the proportion**:
   \[
   \frac{5 \, \text{mL}}{10 \, \text{mg}} = \frac{x \, \text{mL}}{600 \, \text{mg}}
   \]

3. **Solve for \( x \)**:
   Cross-multiply and solve for \( x \):
   \[
   5 \times 600 = 10 \times x
   \]
   \[
   3000 = 10x
   \]
   \[
   x = \frac{3000}{10} = 300 \, \text{mL}
   \]

So, **600 mg of part C would require 300 mL of the mixture**.

Would you like more details or have any questions about this? Here are 5 related questions for further exploration:

1. What is the concentration of part M in this mixture?
2. How would you calculate the amount of M in a specific volume of the mixture?
3. If the amount of C were doubled, how would that affect the mixture volume needed?
4. How would the ratio change if the mixture contained equal parts M and C?
5. What is the total weight of the mixture in 300 mL?

### Tip:
Always check units when setting up ratios to ensure consistency throughout the problem.

Learn how to calculate the volume of a mixture containing specific amounts of substances. This example breaks down the use of ratios and proportions to find how many ml of a mixture containing 6.25mg of M and 10mg of C per 5ml are needed to obtain 600mg of C. Suitable for students in Grades 6-8.

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