If we want to prepare 125 mL of a 1.245 M aqueous KCl solution, how many grams of KCl (rounded with correct number of significant figures) would we need to put in the solution?
.125 L * 1.245 M = ? mol
look up KCl molar mass and multiply by result of 1st calculation
3 sig fig in answer
To calculate the mass of KCl needed to prepare the solution, we can use the formula:
mass = volume x concentration x molar mass
Given:
- Volume (V) = 125 mL
- Concentration (C) = 1.245 M
- Molar mass of KCl = 74.55 g/mol
Let's substitute the values into the formula:
mass = 125 mL x 1.245 M x 74.55 g/mol
Calculating:
mass = 154.828125 g
Rounding to the correct number of significant figures (3 significant figures) gives us:
mass ≈ 155 g
Therefore, we would need approximately 155 grams of KCl to prepare the 125 mL of a 1.245 M aqueous KCl solution.
To determine the mass of KCl needed to prepare the solution, you can use the formula:
mass = volume × concentration × molar mass
First, let's calculate the moles of KCl needed:
moles = volume × concentration
moles = 125 mL × 1.245 M
Now, we need to convert the volume from milliliters (mL) to liters (L):
volume = 125 mL ÷ 1000 mL/L
Plugging in the values:
moles = 0.125 L × 1.245 M
Now, let's get the molar mass of KCl. KCl consists of potassium (K) and chlorine (Cl), and their atomic masses are:
Potassium (K) = 39.10 g/mol
Chlorine (Cl) = 35.45 g/mol
The molar mass of KCl is then:
molar mass = 39.10 g/mol (K) + 35.45 g/mol (Cl)
molar mass = 74.55 g/mol
Finally, we can calculate the mass:
mass = moles × molar mass
mass = (0.125 L × 1.245 M) × 74.55 g/mol
Calculating this gives us the mass of KCl needed to prepare the solution.