Answer to Question #342528 in General Chemistry for rannie

Question

 The boiling point of an aqueous solution is 102.5°C. What is the freezing point of the solution?

Answer 1

1. Find the molality of the solution. for this we use the formula ∆T = i * m * K_b and derive the formula for finding the molality: m = ∆T / (i * K_b).

∆T — change in boiling point = 2.5ºC

i — van't Hoff Factor = 1 assuming a non-electrolyte as the solute (it isn't stated)

m — molality = moles solute/kg solvent

K_b — boiling point constant = 0.51

m = ∆T / ( i * K_b ) = 2,5 / ( 1 * 0,51 ) = 4.9 m

2. Now substitute this value into the freezing point expression and solve for ∆T:

∆T = i * m * K_f

K_f — freezing point constant = 1.86

∆T =imK_f = 1 * 4.9 * 1.86 = 9.114° C

∆T=9.114° C

So, the freezing point of the solution will be -9.114° C

ANSWER: freezing point = -9.114° C