6) How many grams of copper (II) fluoride, CuF2, are needed to make 6.7 liters of a 1.2 M solution?
To determine the number of grams of copper (II) fluoride (CuF2) needed to make a certain molarity solution, we need to use the formula:
Mass (g) = Molarity (M) * Volume (L) * Molar mass (g/mol).
First, let's calculate the molar mass of CuF2:
Copper (Cu) has an atomic mass of 63.55 g/mol.
Fluorine (F) has an atomic mass of 19.00 g/mol.
Since there are two fluorine atoms in CuF2, we multiply the atomic mass of fluorine by 2.
Molar mass of CuF2 = (63.55 g/mol) + (2 * 19.00 g/mol) = 125.55 g/mol.
Now, let's calculate the number of grams of CuF2 needed:
Mass (g) = 1.2 M * 6.7 L * 125.55 g/mol
Mass (g) = 1008.6 g
Therefore, you will need 1008.6 grams of copper (II) fluoride (CuF2) to make 6.7 liters of a 1.2 M solution.
To find out how many grams of copper (II) fluoride (CuF2) are needed to make the given solution, you'll need to follow these steps:
1. Start by finding the molar mass of CuF2, which consists of one copper (Cu) atom and two fluorine (F) atoms. The atomic mass of Cu is 63.55 g/mol, and the atomic mass of F is 18.99 g/mol.
Molar mass of CuF2 = (63.55 g/mol) + 2 * (18.99 g/mol)
2. Calculate the number of moles of CuF2 needed using the molarity (M) and volume (V) of the solution. The molarity is given as 1.2 M, and the volume is given as 6.7 liters.
Moles = Molarity * Volume
3. Convert the moles of CuF2 to grams using the molar mass of CuF2 calculated in step 1.
Grams = Moles * Molar mass
Now, let's put these steps into action:
1. Molar mass of CuF2 = (63.55 g/mol) + 2 * (18.99 g/mol) = 121.53 g/mol
2. Moles = 1.2 M * 6.7 L = 8.04 mol
3. Grams = 8.04 mol * 121.53 g/mol ≈ 977.17 g
Therefore, approximately 977.17 grams of copper (II) fluoride (CuF2) are needed to make 6.7 liters of a 1.2 M solution.
M = mols/L
solve for moles
moles = grams/molar mass
solve for grams.