How many milliliters of an aqueous solution of 0.138 M potassium nitrate is needed to obtain 19.4 grams of the salt?
How many mols is 19.4 grams KNO3?
mols = grams/molar mass
Then mols solution = M x L
Solve for L and convert to mL.
To solve this problem, we need to use the formula:
Molarity (M) = moles of solute / volume of solvent (in liters)
First, let's find the number of moles of potassium nitrate (KNO3) in 19.4 grams using the formula:
moles = mass / molar mass
The molar mass of potassium nitrate (KNO3) is calculated by adding up the atomic masses of the individual elements:
39.1 g/mol (K) + 14.0 g/mol (N) + (16.0 g/mol x 3) (O) = 101.1 g/mol
So, moles = 19.4 g / 101.1 g/mol = 0.192 moles
Now we can use the formula to find the volume of the solution:
0.138 M = 0.192 moles / volume (in liters)
Rearranging the equation to solve for volume:
volume (in liters) = 0.192 moles / 0.138 M = 1.39 liters
However, the question asks for the volume in milliliters. So, we convert the volume from liters to milliliters:
1.39 liters x 1000 mL / 1 liter = 1390 mL
Therefore, you would need 1390 milliliters of the aqueous solution of 0.138 M potassium nitrate to obtain 19.4 grams of the salt.