how many grams of ammonium sulfate are needed to make a .25 M solution at a concentration of 6M?
same process I showed you under Jeff.
To determine how many grams of ammonium sulfate are needed to make a 0.25 M solution at a concentration of 6 M, we need to use the equation:
Molarity = (moles of solute) / (volume of solution in liters)
First, let's calculate the number of moles of solute required for the desired concentration. The formula of ammonium sulfate is (NH4)2SO4, which means it contains 2 ammonium ions (NH4+) and 1 sulfate ion (SO4^2-).
Step 1: Calculate the moles of ammonium sulfate needed for the desired concentration:
Moles of solute = Molarity × Volume of solution in liters
Moles of solute = 0.25 mol/L × 1 L (assuming 1 liter of solution is desired)
Moles of solute = 0.25 mol
Step 2: Calculate the grams of ammonium sulfate needed using its molar mass:
The molar mass of ammonium sulfate ((NH4)2SO4) can be calculated by adding the molar masses of its constituent elements:
2(N) + 8(H) + (S) + 4(O)
Molar mass = 2(14.01 g/mol) + 8(1.01 g/mol) + 32.07 g/mol + 4(16.00 g/mol)
Molar mass = 132.14 g/mol
Now, let's calculate the grams of ammonium sulfate needed:
Grams of ammonium sulfate = Moles of solute × Molar mass
Grams of ammonium sulfate = 0.25 mol × 132.14 g/mol
Grams of ammonium sulfate = 33.035 g
Therefore, approximately 33.035 grams of ammonium sulfate are needed to make a 0.25 M solution at a concentration of 6 M.