What would be Moles of solute in 50mL? Mass of solute in 50mL?

Let's clarify the calculations for the moles of solute and the mass of solute in the 50 mL of the new stock solution.

### 1. **Moles of Solute in 50 mL**

Given:
- Molarity of the new stock solution, \( C_2 = 6.0 \times 10^{-5} \, \text{M} \)
- Volume of the solution, \( V = 50 \, \text{mL} = 0.050 \, \text{L} \)

The formula to calculate moles is:

\[
\text{Moles of solute} = C_2 \times V
\]

Substituting the values:

\[
\text{Moles of solute} = (6.0 \times 10^{-5} \, \text{M}) \times (0.050 \, \text{L}) = 3.0 \times 10^{-6} \, \text{moles}
\]

### 2. **Mass of Solute in 50 mL**

Given:
- Moles of solute calculated above: \( 3.0 \times 10^{-6} \, \text{moles} \)
- Molecular weight of Allura Red, \( MW = 496.42 \, \text{g/mol} \)

The formula to calculate mass is:

\[
\text{Mass} = \text{Moles} \times \text{Molecular weight}
\]

Substituting the values:

\[
\text{Mass} = (3.0 \times 10^{-6} \, \text{moles}) \times (496.42 \, \text{g/mol}) = 1.48926 \times 10^{-3} \, \text{g} = 1.49 \, \text{mg}
\]

### Summary:
- **Moles of Solute in 50 mL**: \( 3.0 \times 10^{-6} \, \text{moles} \)
- **Mass of Solute in 50 mL**: \( 1.49 \, \text{mg} \)

These values represent the amount of solute present in the 50 mL of your prepared stock solution.

Do you need any more details or have further questions?

### Related Questions
1. How would the moles and mass change if you used a different volume of the initial solution?
2. What would be the mass of the solute in 100 mL of the same stock solution?
3. How does temperature affect the molarity of a solution?
4. What are the implications of a lower molecular weight on the mass required for the same molarity?
5. How could you verify the concentration of your stock solution experimentally?

### Tip
Always make sure to convert volumes to liters when calculating moles from molarity to avoid errors.

Learn how to calculate the moles of solute and the mass of solute in a 50 mL solution using molarity and molecular weight. This problem involves advanced concepts in chemistry and is suitable for advanced high school students and beyond.

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