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A solid hydrate with a mass of 2.125 g was heated to drive off the water. A solid anhydrous residue remained, who’s mass was 1.044 g. Calculate the percent water in the hydrate. If the anhydrous residue has a formula weight of 104 g, how many water molecules are present in each molecule of anhydrous salt?

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The percent water in the hydrate would be 50.87% and anhydrous salts usually have no water molecules.

Mass of hydrated solid = 2.125 g

Mass of anhydrous solid = 1.044 g

Mass of water = mass of hydrated solid - mass of anhydrous solid

                        = 2.125 - 1.044

                          = 1.081 g

Percentage water in the hydrated solid = mass of water/mass of hydrated solid x 100%

                     = 1.081/2.125 x 100

                         = 50.87%

To answer the second question, anhydrous salt molecules have no water in them. Hence, irrespective of the formula weight, an anhydrous molecule of the salt would have zero water molecule.

More on hydrated molecules can be found here: https://brainly.com/question/13079009

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