To find out how many grams of copper (II) fluoride (CuF2) are needed to prepare 3.75 liters of a 2.4 M solution, follow these steps:
1. Understand the given values:
- Volume of the solution = 3.75 liters
- Molarity (concentration) of the solution = 2.4 M (which is 2.4 moles/liter)
- Molar mass of CuF2 = 101.53 g/mol
2. Calculate the number of moles of CuF2 required:
- Moles of solute (CuF2) = Molarity × Volume of the solution
- Moles of CuF2 = 2.4 M × 3.75 L
- Moles of CuF2 = 9.0 moles
3. Convert moles of CuF2 to grams:
- Grams of CuF2 = Moles of CuF2 × Molar mass of CuF2
- Grams of CuF2 = 9.0 moles × 101.53 g/mol
- Grams of CuF2 = 913.77 grams
Therefore, you would need 913.77 grams of copper (II) fluoride (CuF2) to make 3.75 liters of a 2.4 M solution.
