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matrenka [14]
3 years ago
6

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.)
Chemistry
1 answer:
Andre45 [30]3 years ago
7 0
<span>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</span>
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Please help Please help
alexdok [17]
Follow Avogadro’s Number
1 mole = 6.02 x 10^23
So we can do it
4.77x10^25/6.02x10^23 = 79.2 mole
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