You start with 2.5L of a KNO3 stock solution and wish to prepare 10.0 L of 1.5 M KNO3. What percentage (m/v) would the potassium nitrate stock solution need to be if you were to use it all? (MW of KNO3: 101g/mole)
L1 x M1 = L2 x M2
2.5 x M1 = 10.0 x 1.5
M1 = ? = 10*1.5/2.5 = 6.0 M
Now convert 6.0M to % w/v.
M = mols/L = 6.0 mols/L.
1 mol = 101 g, then 6 mols must be 6 mol x 101 g/mol = ?
To determine the percentage (m/v) of the potassium nitrate stock solution, we need to calculate the mass of potassium nitrate required to prepare the desired 10.0 L of 1.5 M KNO3 solution.
Step 1: Calculate the moles of KNO3 required:
Molarity (M) = moles of solute / volume of solvent (in liters)
1.5 M = moles of KNO3 / 10.0 L
moles of KNO3 = 1.5 M × 10.0 L = 15 moles
Step 2: Calculate the mass of potassium nitrate:
Mass (g) = moles × molar mass
Mass (g) = 15 moles × 101 g/mole = 1515 g
Step 3: Calculate the percentage (m/v):
Percentage (m/v) = (mass of solute (g) / volume of solution (mL)) × 100
We want the final volume in liters, so we'll need to convert the 15.0 L to mL.
Percentage (m/v) = (1515 g / (10,000 mL × 1.5)) × 100
Percentage (m/v) = 10.1% (rounded to one decimal place)
Therefore, the potassium nitrate stock solution would need to be approximately 10.1% (m/v) in order to prepare the desired 10.0 L of 1.5 M KNO3 solution.
To find the percentage (m/v) of the potassium nitrate (KNO3) stock solution, you'll need to calculate the mass of KNO3 required to make 10.0 L of a 1.5 M solution. Here's how you can do it step by step:
Step 1: Calculate the molar mass of KNO3.
The molar mass of KNO3 is 101 g/mole, as given in the question.
Step 2: Calculate the number of moles of KNO3 required for 10.0 L of a 1.5 M solution.
To do this, use the equation: moles = molarity x volume
moles = 1.5 mol/L x 10.0 L
moles = 15.0 mol
Step 3: Calculate the mass of KNO3 required.
Mass = moles x molar mass
Mass = 15.0 mol x 101 g/mol
Mass = 1515 g
Step 4: Calculate the percentage (m/v).
Percentage (m/v) = (mass/volume) x 100
Percentage (m/v) = (1515 g/2500 mL) x 100
Percentage (m/v) = 60.6 %
Therefore, to prepare 10.0 L of 1.5 M KNO3 solution, the potassium nitrate stock solution would need to be approximately 60.6% (m/v) if you were to use it all.