What mass of MgSO4 · 7H2O is required to prepare 700 mL of a 0.342MMgSO4 solution? Answer in units of g.

How many moles do you want? That is M x L = moles.

Then moles = grams/molar mass. Solve for grams.

To find the mass of MgSO4 · 7H2O required to prepare the solution, we need to use the formula:

Mass = Volume × Concentration × Molar mass

1. Calculate the molar mass of MgSO4 · 7H2O:
- MgSO4 molar mass: 24.31 g/mol (Mg) + 32.06 g/mol (S) + 4 × 16.00 g/mol (O) = 120.37 g/mol
- H2O molar mass: 2 × 1.01 g/mol (H) + 16.00 g/mol (O) = 18.02 g/mol
- MgSO4 · 7H2O molar mass: 120.37 g/mol + 7 × 18.02 g/mol = 246.47 g/mol

2. Plug the given values into the formula:
Volume = 700 mL = 0.700 L (convert mL to L)
Concentration = 0.342 M (given)
Molar mass = 246.47 g/mol (calculated in step 1)

Mass = 0.700 L × 0.342 mol/L × 246.47 g/mol

3. Calculate the mass:
Mass = 57.7511 g

Therefore, you would require approximately 57.7511 grams of MgSO4 · 7H2O to prepare 700 mL of a 0.342 M MgSO4 solution.