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HW1_25

# HW1_25 - solution Following the format of earlier problems...

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Homework due 1/25 The antifreeze in a car is a 50% by weight solution of ethylene glycol, C 2 H 6 O 2, in water. Calculate the boiling point and freezing point of this solution. (For H 2 O: K b = 0.52°C/m K f = 1.86°C/m) This is a typical colligative property problem in that you are asked to determine the effect of a solute on a property of the solvent. In this case, it is the freezing and boiling points of the solvent that need to be determined. Since both of these changes involve the molality of the solution, one must first determine the molality ( moles of solute/Kg solvent ) of the solution. This molality is then used to determine the changes in melting point and freezing point of the solution. Since only the concentration of the solution is given, one must first assume some amount of
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Unformatted text preview: solution. Following the format of earlier problems, I will assume 100 g of solution, which is the denominator of the concentration unit given. In 100 grams of solution there are 50 grams of C 2 H 6 O 2 and 50 g of H 2 O. 50 g H 2 O = 0.050 Kg H 2 O. 1 mole C 2 H 6 O 2 # moles C 2 H 6 O 2 = 50 g C 2 H 6 O 2 x --------------------------- = 0.806 mole C 2 H 6 O 2 62 g C 2 H 6 O 2 # moles C 2 H 6 O 2 0.806 mole C 2 H 6 O 2 molality = m = -------------------------- = --------------------------------- = 16.1 m # Kg H 2 O 0.050 Kg H 2 O ∆T b = K b m = (0.52 °C/m)(16.1 m) = 8.4°C BP sol’n = 100°C + ∆T b = 100°C + 8.4°C = 108.4°C ∆T f = K f m = (1.86°C/m)(16.1 m) = 30.0°C FP sol’n = 0°C - ∆T f = 0°C - 30°C = - 30°C...
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