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Calculate the percent by mass of water in copper(II) sulfate pentahydrate. (CuSO4
5H20) (Round to three sig figs)

Sagot :

Answer: 31.9%

Explanation: sorry if the explanation is not that good because I am doing a assignment... but the answer is right

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The percent by mass of water in copper(II) sulfate pentahydrate. CuSO4 5H20 will be 56.60% in the given complex compound.

What is a complex compound?

Complex compounds are the types of the compound in inorganic chemistry that are attached to one central atom and others are the ligands attached to it the bond between them is made due to the formation of a coordinate bond

In copper(II) sulfate pentahydrate copper will be the central atom and hydrate molecules are ligands.

The percent by mass of water = [mass of H20 × 5 /mass of CuSO4 ] × 100

mass of  CuSO4 = 149

mass of H20 × 5 = 90

substituting the value

The percent by mass of water = 90 / 149 × 100

The percentt by mass of water = 56.60%

Therefore, percent by mass of water in copper(II) sulfate pentahydrate. (CuSO45H20) will be 56.60% in the given complex compound.

learn more about complex compound , here :

https://brainly.com/question/15210033

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