To determine how many grams of potassium metabisulfite should be used, we can apply the formula for concentration:

$\text{Concentration} (\%) = \frac{\text{Mass of solute}}{\text{Volume of solution}} \times 100$

Here:

• The concentration is $0.55\%$.
• The volume of solution is $500 \, \text{ml}$.
• The mass of solute (potassium metabisulfite) is unknown.

Rearranging the formula to solve for mass of solute:

$\text{Mass of solute} = \frac{\text{Concentration} \times \text{Volume}}{100}$

Substitute the values:

$\text{Mass of solute} = \frac{0.55 \times 500}{100} = \frac{275}{100} = 2.75 \, \text{grams}$

So, 2.75 grams of potassium metabisulfite should be used.

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Would you like a more detailed explanation? Let me know if you need any further clarifications!

Here are some related questions you might find interesting:

1. How would you prepare a 1% solution of potassium metabisulfite with the same volume?
2. What is the significance of the percentage concentration in solution preparation?
3. How would you calculate the molar mass of potassium metabisulfite?
4. If the solution was prepared with a different solvent, would the calculation change? Why or why not?
5. How could the preparation of a 0.5% solution of potassium metabisulfite differ from a 0.55% solution?

Tip: Always make sure to double-check the unit conversions (e.g., from mL to L or grams to mg) if they are involved in your calculations.