This preview shows page 1. Sign up to view the full content.
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) = 8.4C BP soln = 100C + T b = 100C + 8.4C = 108.4C T f = K f m = (1.86C/m)(16.1 m) = 30.0C FP soln = 0C  T f = 0C  30C =  30C...
View
Full
Document
 Spring '08
 Martin

Click to edit the document details