Question

Calculate the molarity of a solution that contains 183.51 grams of lead (II) bromide in 500.0 mL of the solution

Answer 1

Answer: 1M

Step-by-step explanation:

Molarity = mols/L

moles of lead bromide: 183.51/ 367.0 = 0.5 mol

500 ml/ 1000 mL = .5L

.5 mol / .5 L = 1 mol/L = 1M