How would you prepare 250.0 ml of a 0.6 M aluminum sulfate solution
I would add to xxxgrams of Aluminum sulfate water up to the 250ml mark, stir, close, and label.
xxx gram=molmassAlsufate*.250*.6
To prepare a 0.6 M aluminum sulfate solution, you will need to determine the amount of aluminum sulfate needed in grams. Here are the steps to calculate and prepare a 250.0 ml solution:
Step 1: Find the molar mass of aluminum sulfate (Al₂(SO₄)₃).
- The molar mass of aluminum (Al) is 26.98 g/mol.
- The molar mass of sulfur (S) is 32.06 g/mol.
- The molar mass of oxygen (O) is 16.00 g/mol.
- The molar mass of aluminum sulfate can now be calculated by:
Molar mass of Al₂(SO₄)₃ = (2 x molar mass of Al) + (3 x molar mass of S) + (12 x molar mass of O)
Step 2: Calculate the amount of aluminum sulfate needed in grams.
- Molarity (M) is defined as moles of solute per liter of solution. In this case, the desired molarity is 0.6 M.
- Moles of aluminum sulfate = Molarity x Volume (in liters)
moles = 0.6 M x 0.250 L = 0.150 moles
- Mass (in grams) of aluminum sulfate = moles x molar mass
Step 3: Convert the mass of aluminum sulfate to the required volume (in milliliters).
- Determine the molarity of the solution desired (0.6 M).
- Convert the moles of aluminum sulfate to milliliters using the following formula:
Volume (ml) = (Mass of aluminum sulfate / Molar mass of aluminum sulfate) x 1000
By following these steps, you can determine the amount of aluminum sulfate needed to prepare a 250.0 ml solution with a concentration of 0.6 M.