How many grams of Kl are required to prepare 500 mL of 0.04 M Kl?
How many mols do you want?
mols KI = M x L = ?
Then mols = g/molar mass
You know mols and molar mass, solve for grams KI
To calculate the number of grams of Kl (potassium chloride) required to prepare a given solution, we need to use the equation:
moles = concentration (M) × volume (L)
Given:
Concentration (M) = 0.04 M
Volume (L) = 500 mL = 0.5 L
First, let's rearrange the equation to solve for moles:
moles = concentration × volume
Now, plug in the given values:
moles = 0.04 M × 0.5 L
moles = 0.02 moles
The number of moles of Kl required is 0.02 moles.
Now, to calculate the mass of Kl (potassium chloride) required, we need to know the molar mass of Kl. The molar mass of Kl is about 74.55 g/mol.
mass (grams) = moles × molar mass
mass (grams) = 0.02 moles × 74.55 g/mol
mass (grams) = 1.49 grams (approximately)
Therefore, approximately 1.49 grams of Kl are required to prepare 500 mL of 0.04 M Kl solution.