Answer :
Final answer:
When heated at 110°C, copper (II) sulfate pentahydrate loses four of its five water molecules of hydration. 10 g of this material contains 3.61 g of water.
Explanation:
When copper (II) sulfate pentahydrate is heated at 110°C, it loses four out of its five water molecules of hydration, leaving behind one water molecule. The formula for copper (II) sulfate pentahydrate is CuSO4·5H2O, and the formula for copper (II) sulfate monohydrate is CuSO4·H2O. To determine how much water remains in 10 g of the material, we can use the molar mass of copper (II) sulfate pentahydrate to find the number of moles, and then convert to the number of moles of water molecules lost. Finally, we can convert the moles of water lost to grams using the molar mass of water (18.015 g/mol).
First, let's calculate the molecular mass of copper (II) sulfate pentahydrate:
(1x Cu) + (1x S) + (4x O) + (10x H2O) = 1x 63.55 g/mol + 1x 32.06 g/mol + 4x 16.00 g/mol + 10x 18.015 g/mol = 249.68 g/mol
Next, let's calculate the number of moles of copper (II) sulfate pentahydrate:
10 g / (249.68 g/mol) = 0.0401 mol
Since there are five water molecules of hydration in one molecule of copper (II) sulfate pentahydrate, we multiply the number of moles of copper (II) sulfate pentahydrate by 5 to find the moles of water lost:
0.0401 mol * 5 = 0.2005 mol
Finally, we convert the moles of water lost to grams:
0.2005 mol * 18.015 g/mol = 3.61 g
Therefore, 10 g of copper (II) sulfate pentahydrate contains 3.61 g of water.
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