You need to make an aqueous solution of 0.233 M magnesium nitrate for an experiment in lab, using a 125 mL volumetric flask. How much solid magnesium nitrate should you add? (fill in blank)
__ grams
See your former post and work backwards to that.
How many mols do you need of Mg(NO3)2?
That's mols = M x L = ?
Then mols = grams/molar mass. Solve for grams Mg(NO3)2.
Post your work if you get stuck.
To determine the amount of solid magnesium nitrate needed, we can use the equation:
Molarity (M) = moles of solute (mol) / volume of solution (L)
We have the molarity (0.233 M) and the volume of solution (125 mL = 0.125 L), so we can rearrange the equation to solve for moles of solute:
moles of solute (mol) = Molarity (M) x volume of solution (L)
Substituting the known values into the equation:
moles of solute (mol) = 0.233 M x 0.125 L
Now, we need to determine the molar mass of magnesium nitrate (Mg(NO3)2) in order to convert moles to grams. The molar mass can be calculated by adding up the atomic masses of each element:
Mg: 24.31 g/mol
N: 14.01 g/mol
O: 16.00 g/mol (there are three oxygen atoms)
Molar mass of Mg(NO3)2 = (24.31 g/mol) + (2 * (14.01 g/mol)) + (6 * (16.00 g/mol))
Now we can convert moles to grams using the molar mass:
moles of solute (mol) x molar mass (g/mol) = grams of solute (g)
Substituting the known values into the equation:
(0.233 M x 0.125 L) x [(24.31 g/mol) + (2 * (14.01 g/mol)) + (6 * (16.00 g/mol))]
Simply evaluate the equation and calculate:
grams of solute (g) = ___________________
By plugging in the given values and performing the calculations, you can determine the amount of solid magnesium nitrate that needs to be added in grams.