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Compute how many grams of aluminum sulfate are required to pre pare 5.30x10 2ml of a 0.2450 M aluminum sulfate solution.

5.30 x 10 2 ml divided by 1000 ml = 0.530 L

M 0.2450 x 0.530 = 0130 mols of Al2 (SO4)3

0.130 mol Al2 (SO4)3 x 342.17 g Al2 (SO4)3 = 44.5 grams of Al2 (SO4)3

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Category:- Chemistry
Reference No.:- M91194953

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