what mass (in grams) of solid potassium permanaganate is required to prepare 250mL of a 0.02 M of potassium permanaganate solution?
answered below
To determine the mass of solid potassium permanganate required to prepare the given solution, you need to use the equation:
Molarity (M) = (moles of solute) / (volume of solution in liters)
First, rearrange the equation to solve for moles of solute:
moles of solute = Molarity (M) * volume of solution (L)
Next, convert the volume of solution from milliliters (mL) to liters (L):
250 mL = 250 / 1000 = 0.250 L
Now, substitute the given values into the equation:
moles of solute = 0.02 M * 0.250 L
Calculate the moles of solute:
moles of solute = 0.005 moles
The molar mass of potassium permanganate (KMnO4) is:
K: 39.10 g/mol
Mn: 54.94 g/mol
O: 16.00 g/mol (4 atoms)
Total molar mass = 39.10 + 54.94 + (16.00 * 4) = 158.04 g/mol
Finally, calculate the mass of solid potassium permanganate:
mass of solid = moles of solute * molar mass
mass of solid = 0.005 moles * 158.04 g/mol
Therefore, the mass of solid potassium permanganate required to prepare 250 mL of a 0.02 M solution is approximately 0.79 grams.