To solve this problem, we need to calculate how much of the 20.5% w/w solution of methanol in acetone is required to get 65.00 g of methanol.

Steps:

1. Understand the percent concentration: A 20.5% w/w solution means that for every 100 g of the solution, there are 20.5 g of methanol.

2. Set up the proportion: We are trying to find the mass of the solution that contains 65.00 g of methanol. Let the mass of the solution required be denoted by $x$.

$\text{mass of methanol} = 20.5\% \times \text{mass of solution}$

$65.00 \, \text{g} = 0.205 \times x$

3. Solve for $x$: To find the mass of the solution, rearrange the equation:

$x = \frac{65.00 \, \text{g}}{0.205}$

Calculation:

$x = \frac{65.00}{0.205} = 317.07 \, \text{g}$

Thus, the student would need 317.07 g of the solution.

Check availability:

The student has 0.20 kg of solution available, which is 200.0 g. Since 317.07 g is greater than 200.0 g, there is not enough solution available.

Final Answer:

No solution (since 317.07 g exceeds the available 200.0 g).

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Would you like more details or have any questions about this solution? Here are some related questions for you to explore:

1. How would you calculate the amount of solute in a solution with a different concentration (e.g., 30%)?
2. What is the molarity of a solution if you know the mass of solute and volume of the solution?
3. How can you convert from a weight percent to a molarity or molality?
4. If the student had enough solution, how would you determine the amount of acetone in the solution?
5. How does temperature affect the concentration of solutions?

Tip: Always pay attention to significant figures, especially when using percentages or converting between