Respuesta :
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 °C
This is the equation if you were to do it with different numbers.
The study of chemicals and bonds is called chemistry. There are different types of elements and these are rare metals and nonmetals.
The correct answer is 100.99C
What is the freezing point?
The temperature at which the liquid starts freezing too solid is called the freezing point.
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 ethylene glycol is 62.07 s/mol.
[tex]\frac{\frac{(373.5 g C2H6O2)}{(62.07 g/mol C2H6O2)}}{(3099 g H2O)} * (1000 g H2O) = 1.9417 m[/tex]
[tex](1.9417 m) * (1.86C/m) = 3.61C change[/tex]
[tex]0C - 3.61C = -3.61C \\(1.9417 m) * (0.512C/m) = 0.994C change \\100C + 0.994 C = 100.99 C[/tex]
Hence, the correct answer is 100.99.
For more information about the freezing point, refer to the link:-
https://brainly.in/question/9531558
