What mass of MgSO4 · 7 H2O is required to
prepare 300 mL of a 0.511 M MgSO4 solution?
How many moles MgSO4.7H2O do you need? That is M x L = moles.
grams = moles/molar mass. Solve for grams.
To determine the mass of MgSO4 · 7 H2O required to prepare a certain volume and concentration of solution, you'll need to know the molar mass of MgSO4 · 7 H2O and follow these steps:
Step 1: Write down the given information:
- Volume of solution = 300 mL = 0.300 L
- Concentration of MgSO4 solution = 0.511 M
Step 2: Determine the number of moles of MgSO4 required:
Use the formula: Moles = Concentration × Volume
Moles of MgSO4 = 0.511 M × 0.300 L
Step 3: Calculate the molar mass of MgSO4 · 7 H2O:
The molar mass of anhydrous MgSO4 is 120.37 g/mol, and each water molecule (H2O) has a molar mass of 18.02 g/mol. Since there are 7 water molecules, the molar mass of MgSO4 · 7 H2O is:
Molar mass of MgSO4 · 7 H2O = (120.37 g/mol) + (7 × 18.02 g/mol)
Step 4: Calculate the mass of MgSO4 · 7 H2O:
Mass = Moles × Molar mass
Mass of MgSO4 · 7 H2O = (0.511 M × 0.300 L) × [(120.37 g/mol) + (7 × 18.02 g/mol)]
By following these steps and plugging in the appropriate values, you can determine the mass of MgSO4 · 7 H2O required to prepare the given solution.