How many grams of Al(ClO4)3 (Mm=326 g/mol) must be added to 105 g of water to make a solution with a colligative molality of 0.315 m?

Answer 1

What is the definition of molal (m)? It is moles of solute per kg of solvent.

Since m = 0.315, this means we have 0.315 moles Al(ClO₄)₃ in 1 kg of water.

Since we do not have 1 kg of water, but rather we have only 0.105 kg water (105 g), we also don't have 0.315 moles of Al(ClO₄)₃. But it's easy to find out how many moles of Al(ClO₄)₃ we do have.

0.315 mol Al(ClO₄)₃/kg water x 0.105 kg water = 0.033075 moles Al(ClO₄)₃

Now simply convert moles Al(ClO₄)₃ to grams using the molar mass of 326 g/mol....

0.033075 moles x 326 g/mol = 10.78 g = 10.8 g Al(ClO₄)₃ (to 3 significant figures)