how many grams of potassium nitrate are needed to prepare 250.0ml of a 0.03M potassium nitrate solution?
.03 / 4 = .0075 mols of KNO3 needed
.0075 * 101 g/mol = .758 grams
How many mols do you want? That's mols = M x L = 0.03 x 0.250 = 0.0075 mols. Then
0.0075 mols x 101 g/mol= ?
To find the number of grams of potassium nitrate needed to prepare the given solution, you'll need to use the formula:
moles = concentration × volume
Step 1: Convert the given volume from milliliters to liters:
250.0 ml ÷ 1000 = 0.250 L
Step 2: Plug the given values into the formula to calculate the number of moles of potassium nitrate:
moles = 0.03 M × 0.250 L = 0.0075 moles
Step 3: Calculate the molar mass of potassium nitrate (KNO3):
Potassium (K) = 39.10 g/mol
Nitrogen (N) = 14.01 g/mol
Oxygen (O) = 16.00 g/mol
Molar mass of KNO3 = (39.10 g/mol) + (14.01 g/mol × 1) + (16.00 g/mol × 3) = 39.10 g/mol + 14.01 g/mol + 48.00 g/mol = 101.11 g/mol
Step 4: Convert the moles of potassium nitrate to grams using the molar mass:
grams = moles × molar mass = 0.0075 moles × 101.11 g/mol ≈ 0.758 grams
Therefore, approximately 0.758 grams of potassium nitrate are needed to prepare 250.0 ml of a 0.03M potassium nitrate solution.