What mass of MgSO4 · 7H2O is required to
prepare 300 mL of a 0.61 M MgSO4 solution?
Answer in units of g
To calculate the mass of MgSO4 · 7H2O required to prepare the 0.61 M MgSO4 solution, we need to first determine the number of moles of MgSO4 needed and then convert it to mass using the molar mass of MgSO4 · 7H2O.
1. Start by using the formula for molarity:
Molarity (M) = moles of solute / volume of solution (in liters)
2. Rearrange the formula to solve for moles of solute:
Moles of solute = Molarity × volume of solution (in liters)
3. Convert the given volume of the solution from milliliters (mL) to liters (L):
300 mL = 300/1000 L = 0.3 L
4. Substitute the values into the formula:
Moles of MgSO4 = 0.61 M × 0.3 L
5. Calculate the moles of MgSO4:
Moles of MgSO4 = 0.183 mol
6. The molar mass of MgSO4 · 7H2O can be calculated using the atomic masses of Mg, S, O, and H:
Mg: 24.305 g/mol
S: 32.06 g/mol
O: 16.00 g/mol
H: 1.008 g/mol
MgSO4 · 7H2O = (24.305 g/mol × 1) + (32.06 g/mol × 1) + (16.00 g/mol × 4) + (1.008 g/mol × 14)
MgSO4 · 7H2O = 246.47 g/mol
7. Finally, calculate the mass of MgSO4 · 7H2O using the moles of MgSO4 calculated earlier:
Mass of MgSO4 · 7H2O = Moles of MgSO4 × molar mass of MgSO4 · 7H2O
Mass of MgSO4 · 7H2O = 0.183 mol × 246.47 g/mol
Therefore, the mass of MgSO4 · 7H2O required to prepare 300 mL of a 0.61 M MgSO4 solution is approximately 45.10 grams.
molarity=mass/formula mass * 1/volume
solveing for mass
mass=molarity*formulamass*volumeinLiters