Calculate how many grams of FeSO4.7 H2O (MW = 278.02 g/mol) will be needed to prepare 1 mL of a 0.2 M FeSO4 solution?

Is that .0304g?

I don't think so.

mols you need = M x L = ?
Then grams = mols x molar mass = ?

To calculate the number of grams of FeSO4.7H2O needed to prepare 1 mL of a 0.2 M FeSO4 solution, you will need to determine the number of moles of FeSO4 required first and then convert it to grams using the molar mass.

First, let's determine the number of moles of FeSO4 needed:
Molarity (M) is defined as moles of solute per liter of solution. Since we want to prepare 1 mL (0.001 L) of a 0.2 M solution, we can use the following formula:

Moles = Molarity × Volume
Moles = 0.2 mol/L × 0.001 L
Moles = 2 × 10^-4 moles

Next, we need to convert the moles of FeSO4 to grams using the molar mass:
The molecular weight (MW) of FeSO4.7H2O is 278.02 g/mol. However, we need to consider the water of crystallization (7H2O) when calculating the effective molar mass of the compound.

Effective molar mass of FeSO4.7H2O = MW of FeSO4.7H2O / 1 mole of FeSO4.7H2O
Effective molar mass of FeSO4.7H2O = 278.02 g/mol / 1
Effective molar mass of FeSO4.7H2O = 278.02 g/mol

Finally, we can calculate the mass of FeSO4.7H2O needed:
Mass (g) = Moles × Effective molar mass
Mass (g) = 2 × 10^-4 moles × 278.02 g/mol

After performing the calculation, the answer is 0.0556 grams of FeSO4.7H2O needed to prepare 1 mL of a 0.2 M FeSO4 solution.

So, the correct answer is not 0.0304g, but rather 0.0556g.