How many grams of KI must be added to a 250-ml volumetric flask to prepare a water solution that is 0.100 M in potassium iodide?
How many mols KI do you want? That's mols = M x L = ?
Then mols = grams/molar mass. You know mols and molar mass, solve for grams.
To determine how many grams of KI must be added to prepare a 0.100 M potassium iodide solution in a 250-ml volumetric flask, we start by understanding the steps involved:
Step 1: Determine the molar mass of KI.
Step 2: Use the molar mass to convert the desired molarity (M) to moles.
Step 3: Convert moles to grams using the stoichiometry of KI.
Step 1: Determine the molar mass of KI.
The molar mass of KI can be calculated by adding the atomic masses of potassium (K) and iodine (I).
Molar mass of KI = atomic mass of K + atomic mass of I
Looking up the atomic masses:
Atomic mass of K = 39.10 g/mol
Atomic mass of I = 126.90 g/mol
Molar mass of KI = 39.10 g/mol + 126.90 g/mol = 166.00 g/mol
Step 2: Convert the desired molarity to moles.
Molarity (M) is defined as moles of solute per liter of solution.
0.100 M potassium iodide solution means 0.100 moles of KI in 1 liter of solution (1 L = 1000 ml).
Given:
Molarity (M) = 0.100 M
Volume (V) = 250 ml = 0.250 L
Moles (n) = Molarity (M) x Volume (V) = 0.100 mol/L x 0.250 L = 0.025 mol
Step 3: Convert moles to grams using the stoichiometry of KI.
From Step 2, we know that there are 0.025 moles of KI needed.
Using the molar mass of KI calculated in Step 1, we can calculate the grams of KI:
Grams (g) = Moles (n) x Molar mass (M)
Grams (g) = 0.025 mol x 166.00 g/mol = 4.15 g
Therefore, to prepare a 0.100 M potassium iodide solution in a 250-ml volumetric flask, you must add 4.15 grams of KI.