To find the concentration in % m/v (mass/volume percent), we use the following relationship:

$\text{Concentration} \, (\% m/v) = \frac{\text{mass of solute (g)}}{\text{volume of solution (mL)}} \times 100$

Given:

• Molarity (M) = 0.677 mol/L
• Molar mass (MM) of acetone = 58.08 g/mol
• 1 L = 1000 mL

Step 1: Calculate the mass of acetone in 1 liter of solution.

The molarity tells us that there are 0.677 moles of acetone per liter of solution.

$\text{Mass of acetone (g)} = \text{Molarity (mol/L)} \times \text{Molar mass (g/mol)} = 0.677 \, \text{mol/L} \times 58.08 \, \text{g/mol} = 39.33 \, \text{g}$

Step 2: Calculate the concentration in % m/v.

Now that we know there are 39.33 g of acetone in 1 L (1000 mL) of solution:

$\text{Concentration} \, (\% m/v) = \frac{39.33 \, \text{g}}{1000 \, \text{mL}} \times 100 = 3.933\%$

Final Answer:

The concentration of the solution is 3.933% m/v.

Would you like more details on any step?

Here are 5 related questions for practice:

1. How would the % m/v change if the molarity were doubled?
2. What is the molarity of a solution with a concentration of 5% m/v?
3. How do you convert between % m/v and molarity for any solute?
4. If the volume of solution is reduced by half, what happens to the % m/v concentration?
5. How does the density of the solution affect % m/v?

Tip: Always double-check your units to ensure you're calculating mass/volume percent correctly!