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An ethylene glycol solution contains 16.2 g of ethylene glycol (c2h6o2) in 85.4 ml of water. calculate the boiling point of the solution. calculate the freezing point of the solution. (assume a density of 1.00 g/ml for water.)

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Answer: (16.2 g C2H6O2) / (62.0678 g C2H6O2/mol) / (0.0982 kg) = 3.9704 mol/kg = 3.9704 m a.) (3.9704 m) x (1.86 °C/m) = 7.38 °C change 0.00°C - 7.38 °C = - 7.38 °C b.) (3.9704 m) x (0.512 °C/m) = 2.03 °C change 100.00°C + 2.03 °C = 102.03 °C

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