Question

What is the molarity of 122.5 g of AlCl3 in 1.0 L of solution? (MM = 133 g/mol)

Answer 1

Answer: Molarity is defined as moles of solute per liter of solution. So, find the moles of solute and divide by the liters of solution.
molar mass AlCl3 = 133g/mole
moles AlCl3 = 127 g x 1 mole/133 g = 0.955 moles
liters of solution = 400 ml x 1 liter/1000 ml = 0.400 liters
Molarity = 0.955 moles/0.400 liters = 2.39 M
Explain: I looked it up on wyzant.com