# Calculate freezing point depression and boiling point elevation

by flemonster
Tags: boiling, depression, elevation, freezing, point
 P: 11 1. The problem statement, all variables and given/known data Calculate new freezing point and boiling point when 0.47 mol ethylene glycol and 0.14 mol KBr is added to 150g H2O. Express your answer using one decimal place. 2. Relevant equations molality = moles solute/kg solvent ΔTf=i*Kf*m ΔTb=i*Kb*m New bp = ΔTb + normal bp New fp = normal fp -ΔTf ΔTf = change in freezing point ΔTb = change in boiling point m = molality of solution Kb = boiling point constant Kf = freezing point constant i = number of solute particles 3. The attempt at a solution Molality of solution = (0.14 mol KBr +0.47 mol ethylene glycol)/(0.150 kg H2O) = 4.1 m Kf H2O = 1.86°C/m (from table provided in the problem) i = 3 (ethylene glycol is a covalently bonded molecule but potassium bromide dissociates into its constituent ions) ΔTf = 3*1.86*4.1 = 22.9 New fp = 0 - 22.9 = -22.9°C Kb H2O = 0.51°C/m (from table provided in the problem) ΔTb = 3*0.51*4.1 = 6.3 New bp = 100 + 6.3 = 106.3°C EDIT: I put the bp into mastering chemistry and it told me that I might have used the wrong number of sig figs or have committed a rounding error. I have no idea what I am doing wrong because I have checked and double checked the calculations. I also put in 106.2 into the answer box and it said the same thing about rounding and sig figs. I'm hesitant to try the fp as I have already been marked down. Any help?
P: 23,599
Sorry for a late answer, somehow I missed your question on the original date.

 Quote by flemonster Molality of solution = (0.14 mol KBr +0.47 mol ethylene glycol)/(0.150 kg H2O) = 4.1 m
 i = 3 (ethylene glycol is a covalently bonded molecule but potassium bromide dissociates into its constituent ions) ΔTf = 3*1.86*4.1 = 22.9
It doesn't work this way. What you did suggests ethylene glycol is dissociated into three ions.

Think about it this way - you need to calculate molality of ALL entities (ions, molecules) present in the solution. You put 0.47 moles of ethylene glycol into solution - it doesn't dissociate any longer. You put 0.14 moles of KBr into solution - it is equivalent to putting 0.14 moles of K+ and 0.14 moles of Br-. Can you sum all these numbers of moles now and use it for molality calculations?
P: 11
 Quote by Borek Sorry for a late answer, somehow I missed your question on the original date. It doesn't work this way. What you did suggests ethylene glycol is dissociated into three ions. Think about it this way - you need to calculate molality of ALL entities (ions, molecules) present in the solution. You put 0.47 moles of ethylene glycol into solution - it doesn't dissociate any longer. You put 0.14 moles of KBr into solution - it is equivalent to putting 0.14 moles of K+ and 0.14 moles of Br-. Can you sum all these numbers of moles now and use it for molality calculations?
Hey Borek,

I got it figured out later but thanks anyways for the help. A late response is better than none at all!

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