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Chemistry 18 Online
OpenStudy (lena772):

How many grams of Aluminum chloride are present in 550.0 mL of an Aluminum chloride solution that is 0.083369 M? I got 20.21g, but that's incorrect.

OpenStudy (surry99):

1) Calculate the number of moles of the AlCl3 solution .083369 mol/l x .550 l = .045 mol 2) Determine the molecular weight of AlCl3 27 + (3 x 35.5)= 133.5 g/mol 3) .045 mol x 133.5 g/mol = 6.00 g

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