Explanation Mass of CuSO 4 ⋅ 5H 2 O is given as 360 g . The molar mass of CuSO 4 ⋅ 5H 2 O is calculated as shown below; Molar mass = 1 × M ( Cu ) + 1 × M ( S ) + 4 × M ( O ) + 5 × M ( H 2 O ) = 1 ( 63.55 g ⋅ mol − 1 ) + 1 ( 32.06 g ⋅ mol − 1 ) + 4 ( 16.00 g ⋅ mol − 1 ) + 5 ( 18.016 g ⋅ mol − 1 ) = 63.55 g ⋅ mol − 1 + 32.06 g ⋅ mol − 1 + 64.00 g ⋅ mol − 1 + 90.08 g ⋅ mol − 1 = 249.69 g ⋅ mol − 1 One mole of CuSO 4 ⋅ 5H 2 O contains five moles of water. This means 249.69 g ⋅ mol − 1 of CuSO 4 ⋅ 5H 2 O contains 90.08 g ⋅ mol − 1 of water. Therefore, the amount of water present in 360 g of CuSO 4 ⋅ 5H 2 O is calculated as follows; m ( H 2 O ) = 360 g CuSO 4 ⋅ 5H 2 O × 1 mol CuSO 4 ⋅ 5H 2 O 249

ACHIEVE/CHEMICAL PRINCIPLES ACCESS 2TERM

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ISBN: 9781319403959

Author: ATKINS

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Chapter F, Problem E.30E

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Mass of anhydrous copper(II) sulfate that will remain after 90 % removal of water from CuSO4⋅5H2O has to be calculated.

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Chapter F, Problem E.29E

Chapter F, Problem E.31E

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ACHIEVE/CHEMICAL PRINCIPLES ACCESS 2TERM

Ch. F - Prob. A.1AST Ch. F - Prob. A.1BST Ch. F - Prob. A.2AST Ch. F - Prob. A.2BST Ch. F - Prob. A.3AST Ch. F - Prob. A.3BST Ch. F - Prob. A.4AST Ch. F - Prob. A.4BST Ch. F - Prob. A.5AST Ch. F - Prob. A.5BST

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Mass of anhydrous copper(II) sulfate that will remain after 90 % removal of water from CuSO 4 ⋅ 5H 2 O has to be calculated.

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Interpretation: Mass of anhydrous copper(II) sulfate that will remain after 90 % removal of water from CuSO4⋅5H2O has to be calculated.

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Chapter F Solutions

Interpretation:

Mass of anhydrous copper(II) sulfate that will remain after 90 % removal of water from CuSO4⋅5H2O has to be calculated.