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4. How many grams of copper (II) fluoride, CuF₂ (MW: 101.53), are needed to make 3.75 liters of a 2.4 M solution?

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GinnyAnswer
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.
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