How many moles of chloride ions are in 250.0 ml of a 0.42 M aluminum chloride solution?

Question

Question

Answer 1

Answer #1

molarity = number of moles / volume of solution in L

0.42 M = number of moles / 0.250 L

number of moles = 0.42 * 0.250 = 0.105 mole

Number of moles = mass of aluminum chloride / molar mass of aluminum chloride

0.105 mole = mass of aluminum chloride / 133.34 g/mol

mass of aluminum chloride = 0.105 mole * 133.34 g/mol = 14.0 g

Therefore, the mass of aluminum chloride = 14.0 g

Answer 2

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