更新1:
a) 0.613 m b) 0.895 m c) 1.39 m d) 0.452 m e) None of the choices listed are correct.
What is the molality (m) of a solution prepared by dissolving 125.00 g of ethylene glycol (C2H6O2) in 2.25 kg of water?
回答 (3)
2016-07-19 4:13 pm
Number of moles of C₂H₆O₂ = (125.00 g) / (62.068 g/mol) = 2.014 mol
Mass of water = 2.25 kg
Molality = (2.014 mol) / (2.25 kg) = 0.895 m
...... The answer is: 0.895 m
Mass of water = 2.25 kg
Molality = (2.014 mol) / (2.25 kg) = 0.895 m
...... The answer is: 0.895 m
2016-07-19 7:59 pm
B
2016-07-19 4:13 pm
mol EG in 125.00g = 125.00/62.068 = 2.0139 mol dissolved in 2.25kg water
molality = 2.0139/2.25
Molality = 0.895m
Choice b) is correct.
molality = 2.0139/2.25
Molality = 0.895m
Choice b) is correct.
收錄日期: 2021-04-18 15:20:55
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