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A 0.755-g sample of hydrated copper(II) sulfate CuSO4 · xH2O was heated carefully until it had changed completely to anhydrous copper(II) sulfate (CuSO4) with a mass of 0.483 g. Determine the value of x. [This number is called the number of waters of hydration of copper(II) sulfate. It specifies the number of water molecules per formula unit of CuSO4 in the hydrated crystal.]

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There are five molecules of the water of crystallization.

Mass of hydrated copper(II) sulfate CuSO4 · xH2O = 0.755 g

Mass of anhydrous copper(II) sulfate CuSO4 = 0.483 g

Molar mass of hydrated copper(II) sulfate CuSO4 · xH2O = (160 + 18x) g/mol

Molar mass of anhydrous copper(II) sulfate CuSO4 =160 g/mol

Number of moles of hydrated copper(II) sulfate CuSO4 · xH2O = Number of moles of anhydrous copper(II) sulfate CuSO4

Therefore;

0.755 g/(160 + 18x) g/mol  = 0.483 g/160 g/mol

0.755 * 160 = 0.483 (160 + 18x)

120.8 = 77.28 * 8.694x

120.8 - 77.28 = 8.694x

x = (120.8 - 77.28)/8.694

x = 5

The formula of the hydrated salt is CuSO4 · 5H2O.

Learn more; https://brainly.com/question/14252791

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