How many grams of Al2(SO4)3 are needed to make 87.5 g of 0.3 m Al2(SO4)3 solution?
Kaplan's solution:
step1). formula weight of aluminum sulfate as 342 g/mol therefore 102.6 grams are required per kilogram of solvent (0.3 x 342 = 102.6).
step2).total mass of a solution produced by dissolving 102.6 g of aluminum sulfate in 1kg i.e., 1000 g, of solvent will then be 1102.6 grams.
step3). We can set up and rearrange a ratio now, to find the quantity of Al2(SO4)3 required for the 87.5 g of solution given in the question
answer: (87.5) (102.6) / 1102.6
Solution problems in general are giving me a headache..... I understand the basic concept that concentration = amount of solute / amount of solvent, but in this example Kaplan uses a ratio of (g solute / g solution) = (g solute / g solution). Is this a ratio that I can use in other concentration problems? If so please expand on how to set it up. thanks!
Kaplan's solution:
step1). formula weight of aluminum sulfate as 342 g/mol therefore 102.6 grams are required per kilogram of solvent (0.3 x 342 = 102.6).
step2).total mass of a solution produced by dissolving 102.6 g of aluminum sulfate in 1kg i.e., 1000 g, of solvent will then be 1102.6 grams.
step3). We can set up and rearrange a ratio now, to find the quantity of Al2(SO4)3 required for the 87.5 g of solution given in the question
answer: (87.5) (102.6) / 1102.6
Solution problems in general are giving me a headache..... I understand the basic concept that concentration = amount of solute / amount of solvent, but in this example Kaplan uses a ratio of (g solute / g solution) = (g solute / g solution). Is this a ratio that I can use in other concentration problems? If so please expand on how to set it up. thanks!
