# Ethylene glycol [CH(OH)CH2(OH)] is a common automobile antifreeze. Calculate the freezing point and boiling point of a solution containing 273 g of ethylene glycol and 1055 g of water.

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###### Question:

Ethylene glycol [CH(OH)CH2(OH)] is a common automobile antifreeze. Calculate the freezing point and boiling point of a solution containing 273 g of ethylene glycol and 1055 g of water.

## Answers

Ethylene glycol (EG), CH2(OH)CH2(OH), is a common automobile antifreeze. It is water soluble and fairly volatile (b.p. 197°C). Calculate the boiling point and freezing point of a solution containing 373.5 g of ethylene glycol in 3099 g water. The molar mass of ethvlene glvcol is 62.07 s/mol. Equation:(373.5 g C2H6O2) / (62.07 g/mol C2H6O2) / (3099 g H2O) x (1000 g H2O) = 1.9417 m (1.9417 m) x (1.86 °C/m) = 3.61 °C change 0°C - 3.61 °C = -3.61 °C (1.9417 m) x (0.512 °C/m) = 0.994 °C change 100°C + 0.994 °C = 100.99 °CThis is the equation if you were to do it with different numbers.

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