HW1_18 - 5 OH 1 mole H 2 O # moles H 2 O = 85 g H 2 O X

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Homework due 1/18 Given a solution that contains 25 g C 2 H 5 OH dissolved in 85 g H 2 O. Calculate a) the % C 2 H 5 OH, b) the mole fraction C 2 H 5 OH, and c) the molality of the solution . # g C 2 H 5 OH a) % C 2 H 5 OH = ------------------------- X 100 % # g solution # g solution = # g C 2 H 5 OH + # g H 2 O = 25 g C 2 H 5 OH + 85 g H 2 O = 110 g solution 25 g C 2 H 5 OH % C 2 H 5 OH = ------------------------ X 100 % = 22.72 % 110 g solution # mole C 2 H 5 OH b) X C 2 H 5 OH = ------------------------------ # moles of solution 1 mole C 2 H 5 OH # moles C 2 H 5 OH = 25 g C 2 H 5 OH X ------------------------ = 0.54 moles C 2 H 5 OH 46 g C 2 H
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Unformatted text preview: 5 OH 1 mole H 2 O # moles H 2 O = 85 g H 2 O X ------------------------ = 4.72 moles H 2 O 18 g H 2 O total moles = # moles C 2 H 5 OH + # moles H 2 O = = 0.54 moles C 2 H 5 OH + 4.72 mole H 2 O = 5.26 mole solution 0.54 mole C 2 H 5 OH X C 2 H 5 OH = ------------------------------ = 0.103 (no units) 5.26 mole solution # moles of C 2 H 5 OH 0.54 mole C 2 H 5 OH c) m = ------------------------------ = -------------------------------- = 6.35 m # Kg H 2 O(solvent) 0.085 Kg H 2 O...
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This note was uploaded on 04/03/2008 for the course CHEM 114 taught by Professor Martin during the Spring '08 term at Moravian.

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