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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 Spring '08
 Martin
 Propylene glycol, Ethylene glycol, Kg H2O Tb

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