Find the volume in liters of 0.452 M manganese(II) sulfate that contains 54.5 g of solute?
To find the volume in liters of a solution, you need to know the concentration of the solution and the mass of the solute. In this case, you have the concentration (0.452 M) and the mass of the solute (54.5 g).
The concentration (C) is expressed in units of moles per liter (mol/L). This tells you the number of moles of solute present in one liter of the solution. In this case, the concentration is 0.452 M, which means there are 0.452 moles of manganese(II) sulfate in one liter of the solution.
The mass of the solute (m) is given as 54.5 g.
To find the volume (V) of the solution, you can use the equation:
V = (m / M) / C
Where:
- V is the volume of the solution in liters
- m is the mass of the solute in grams
- M is the molar mass of the solute in grams per mole
- C is the concentration of the solution in moles per liter
The molar mass of manganese(II) sulfate (MnSO4) can be calculated by adding up the atomic masses of its individual atoms:
- Atomic mass of Mn = 54.94 g/mol
- Atomic mass of S = 32.06 g/mol
- Atomic mass of O = 16.00 g/mol (there are 4 oxygen atoms in MnSO4)
Molar mass of MnSO4 = (54.94 g/mol) + (32.06 g/mol) + (16.00 g/mol x 4) = 54.94 g/mol + 32.06 g/mol + 64.00 g/mol = 150.00 g/mol
Now, substitute the given values into the equation:
V = (54.5 g / 150.00 g/mol) / 0.452 mol/L
Calculate the value:
V = 0.362 L
Therefore, the volume of the 0.452 M manganese(II) sulfate solution that contains 54.5 g of solute is 0.362 liters.
I'd divide 54.5 g by the molar mass to find the number moles of manganese(II) sulfate (MnSO4).
Molarity (M) = moles of solute/ volume in liters