A 5.12 L sample of solution contains 0.230 g of potassium sulfate, K₂SO₄. Determine the concentration of K₂SO₄ in ppm if the density of the solution is 1.30 g/mL.

• A. 34.6 ppm
• B. 28.9 ppm
• C. 22.3 ppm
• D. 43.6 ppm

Answer: (A)

Verified step by step guidance

1. First, calculate the mass of the solution using its volume and density. The volume of the solution is 5.12 L, which is equivalent to 5120 mL. Use the formula: \( \text{mass of solution} = \text{volume} \times \text{density} \).
2. Substitute the given values into the formula: \( \text{mass of solution} = 5120 \text{ mL} \times 1.30 \text{ g/mL} \). This will give you the total mass of the solution in grams.
3. Next, calculate the concentration of potassium sulfate (K2SO4) in parts per million (ppm). The formula for ppm is: \( \text{ppm} = \left( \frac{\text{mass of solute}}{\text{mass of solution}} \right) \times 10^6 \).
4. Substitute the mass of K2SO4 (0.230 g) and the mass of the solution (calculated in step 2) into the ppm formula: \( \text{ppm} = \left( \frac{0.230 \text{ g}}{\text{mass of solution}} \right) \times 10^6 \).
5. Finally, solve the equation to find the concentration of K2SO4 in ppm. This will give you one of the provided answer choices.