How many grams of KMnO4 Is required to prepare 1.0 liter of a solution of 1.5 M KMnO 4?
The molar mass of KMnO4 is 158.034 g/mol.
To prepare 1.0 liter of a solution of 1.5 M KMnO4, we need to calculate the number of moles of KMnO4 required:
1.5 M = 1.5 mol/L
1.5 mol/L x 1.0 L = 1.5 mol
Now, we can calculate the mass of KMnO4 required using the molar mass:
1.5 mol x 158.034 g/mol = 237.051 g
Therefore, 237.051 grams of KMnO4 is required to prepare 1.0 liter of a solution of 1.5 M KMnO4.
To calculate the number of grams of KMnO4 required to prepare a 1.0 liter solution of 1.5 M KMnO4, you will need to use the formula:
Molarity (M) = moles of solute / liters of solution
First, rearrange the formula to solve for moles of solute:
moles of solute = Molarity * liters of solution
Given:
Molarity (M) = 1.5 M
Liters of solution = 1.0 L
Now substitute the values into the formula:
moles of solute = 1.5 M * 1.0 L
moles of solute = 1.5 moles
Next, use the molar mass of KMnO4 to convert moles to grams. The molar mass of KMnO4 is calculated as follows:
K (potassium) = 39.10 g/mol
Mn (manganese) = 54.94 g/mol
O (oxygen) = 16.00 g/mol (4 oxygen atoms in KMnO4)
Total molar mass = (39.10 g/mol) + (54.94 g/mol) + (16.00 g/mol * 4) = 158.03 g/mol
Finally, calculate the grams of KMnO4 needed:
grams = moles * molar mass
grams = 1.5 moles * 158.03 g/mol
grams = 237.045 g
Therefore, approximately 237.045 grams of KMnO4 is required to prepare 1.0 liter of a 1.5 M KMnO4 solution.