Question: How many grams of KCl are needed to prepare 0.5 M solution in 500 mL of water? (2023)
Options:
Correct Answer: 7.45 g
Solution:
Molarity = moles/volume(L). Moles = 0.5 mol/L * 0.5 L = 0.25 mol. Mass = moles * molar mass = 0.25 mol * 74.5 g/mol = 18.625 g.