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Ch.13 - Solutions
All textbooksTro 4th EditionCh.13 - SolutionsProblem 78
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Chapter 13, Problem 78

An ethylene glycol solution contains 21.2 g of ethylene glycol (C2H6O2) in 85.4 mL of water. Determine the freezing point and boiling point of the solution. (Assume a density of 1.00 g/mL for water.)

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1
Calculate the molality of the solution by first determining the moles of ethylene glycol (C_2H_6O_2) using its molar mass.
Convert the volume of water to mass using the given density (1.00 g/mL), then use this mass to calculate the molality of the solution.
Use the freezing point depression formula, \( \Delta T_f = i \cdot K_f \cdot m \), where \( i \) is the van't Hoff factor (1 for ethylene glycol), \( K_f \) is the freezing point depression constant for water, and \( m \) is the molality.
Calculate the new freezing point by subtracting \( \Delta T_f \) from the normal freezing point of water (0°C).
Use the boiling point elevation formula, \( \Delta T_b = i \cdot K_b \cdot m \), where \( K_b \) is the boiling point elevation constant for water, to find the change in boiling point, and add \( \Delta T_b \) to the normal boiling point of water (100°C).
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