Find the change in freeze, molality of solution, # of mols, molecular weight for the problem below

Question

 Details

-A sample, of a white powder found at Post Office, weighing .95g was mixed with 20g of Hexane.

 -Hexane has a freezing pt. of 6oC​ Kf Hexane is 5.2

 -The solution was placed in a freezer; and the solution was observed to freeze at 4.5oC

 

Question 1: What was the change in freeze. Pt. of the soln from solvent​(Atf)

6oC – 4.5oC = ______________

 

Question 2: Find the molality (m) of the soln. using ​Atf = (kf)(m)​​(SW)

 

 

  Question 3 :Find the number of mols using​​ m = mols/Kg solvent or mols = m/Kg

(20 g Hexane equals .02Kg of Hexane)

​​​​​​​mol = m/.02 use m from above!

 

 

Question 4: Determine the Molecular Wt. of the unknown using MW=g/mol (Use .95 for the grams (g) and mol from above. (SW)

 

 

 

**Question 5: Using the calculated MW; propose a suitable CxHyOz formula like C4H10O9

 

 

Answer 1

∆T = imK

∆T = change in temperature = 60º - 4.5º = 55.5º

This is a very large change in freezing point, so I'm wondering if there is an error somewhere in the question.

i = van't Hoff factor. Since this is an unknown, we don't know i, but since it is soluble in hexane we will assume it has a value of 1 (non-electrolyte)

m = molality = moles of solute/kg hexane = ?

K = freezing point constant for hexane = 5.2

(1). 55.5º

(2). 55.5 = (1)(m)(5.2)

m = 55.5/5.2 = 10.7 m

(3). m = moles/kg; 10.7 mol/kg x 0.02 kg = 0.214 moles (note: I believe the formula given in the question is incorrect, i.e. mol = m/0.02 is not correct. It should be mol = m x 0.02 kg)

(4). molar mass = g/mol = 0.95 g/0.214 moles = 4.4 g/mol (This is not possible for a compound of CxHyOz. This goes back to my original comment above about the large decrease in freezing point. Perhaps the question meant to say that the change in freezing point was 4.5ºC? That would make more sense.