Answer to Question #123747 in Chemistry for santana

Question

What is the molarity of a solution that contains 0.350 g of barium fluoride, BaF2, in 0.50 L of solution?

Answer 1

Solution:

Moles of BaF₂ = Mass of BaF₂ / Molar mass of BaF₂

The molar mass of BaF₂ is 175.324 g/mol

Thus:

Moles of BaF₂ = (0.350 g) / (175.324 g/mol) = 0.001996 mol = 0.002 mol

Molarity = Moles of solute / Liter of solution

Molarity of BaF₂ = Moles of BaF₂ / Liter of solution

Thus:

Molarity of BaF₂ = (0.002 mol) / (0.50 L) = 0.004 mol/L = 0.004 M

Answer: The molarity of a solution is 0.004 M.