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Title: How many milliliters of an aqueous solution of 0.172 M iron(III) bromide is needed to obtain 16.6 grams of the? Post by: firehawk347 on Nov 7, 2012 How many milliliters of an aqueous solution of 0.172 M iron(III) bromide is needed to obtain 16.6 grams of the salt ?
Title: How many milliliters of an aqueous solution of 0.172 M iron(III) bromide is needed to obtain 16.6 grams of the? Post by: DaYea on Nov 7, 2012 1molar solution of FeBr3 = 295.557g/mole So, 0.172mola solution of FeBr3 contains 295.557 x 0.172 = 50.8358 g/1000mL. Therefore to get 16.60g of the salt one needs to have (16.60 x 1000)/50.858 =326.54mL.
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