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Boiling Point Elevation and Freezing Point Depression Calculations

The addition of a nonvolatile solute to a solvent causes the boiling point of the solvent to increase and the freezing point of the solvent to decrease. 

Regents Questions-Follow the Regents link to check the answer

Jan 2003-19 What occurs when NaCl(s) is added to water

(1) The boiling point of the solution increases, and the freezing point of the solution decreases.
(2) The boiling point of the solution increases, and the freezing point of the solution increases.
(3) The boiling point of the solution decreases, and the freezing point of the solution decreases.
(4) The boiling point of the solution decreases, and the freezing point of the solution increases.

 

June 2003-23 At standard pressure when NaCl is added to water, the solution will have a

(1) higher freezing point and a lower boiling point than water
(2) higher freezing point and a higher boiling point than water
(3) lower freezing point and a higher boiling point than water
(4) lower freezing point and a lower boiling point than water

 

Jan 2007-19 Compared to a 2.0 M aqueous solution of NaCl at 1 atmosphere, a 3.0 M aqueous solution of NaCl at 1 atmosphere has a
(1) lower boiling point and a higher freezing point
(2) lower boiling point and a lower freezing point
(3) higher boiling point and a higher freezing point
(4) higher boiling point and a lower freezing point

 

Jun 2009-20 Compared to the freezing point and boiling point of water at 1 atmosphere, a solution of a salt and water at 1 atmosphere has a

(1) lower freezing point and a lower boiling point
(2) lower freezing point and a higher boiling point
(3) higher freezing point and a lower boiling point
(4) higher freezing point and a higher boiling point

Advanced Chemistry Calculations

Table 1

The Change in Boiling Point Equation

The Change in Freezing Point Equation

 DTb=imKb

 DTf=imKf
DTb= change in boiling point

i= van't Hoff factor

m= molality (moles solute/ kg solvent)

Kb= boiling point elevation constant (water Kb=+0.52oC/m)

 DTf= change in freezing point

i= van't Hoff factor

m= molality

Kf= freezing point depression constant (water Kf=-1.86oC/m)
Normal Boiling Point of Water = 100.oC Normal Freezing Point of Water = 0.00oC

Table 2

 

i=van't Hoff= the number of particles a formula breaks up into

C6H12O6(s) ==>C6H12O6(aq)    (1 mole of particles)       C6H12O6 =1

NaCl(s) ==> Na+(aq) + Cl-(aq)    (2 moles of particles)      NaCl=2

CaCl2(s) ==> Ca2+(aq) + 2Cl-(aq)   (3 moles of particles)    CaCl2=3

More on the Van't Hoff Factor

Table 3

For every mole of particles added to1 kg of pure water the freezing point decreases by 1.86oC and the boiling point increases by 0.52oC

 

 

Looking at the equation, the greater the concentration (molality) the greater the change.

 DT=imK

 

Sample Calculations

Determine the boiling point of a solution made from 2kgs of water and 1 mol of C6H12O6.

 DTb=imKb

i=1 (see table 2);     m= mol/kg= 1 mol/2kg= 0.5mol/kg;       Kb=+0.52oC/m (see table 1)

 

DTb= (1)(0.5mol/kg)(+0.52oC/m) =+0.26oC

Boiling point= 100.oC +0.26oC=100.26oC

 

Determine the freezing point of a solution made from 3kgs of water and 2.5mol of CaCl2.

 DTf=imKf

i=3 (see table 2);     m= mol/kg= 2.5 mol/3kg= 0.83mol/kg;       Kf=-1.86oC/m (see table 1)

 

DTf=3(0.83mol/kg)(-1.86oC/m)= -4.63

Freezing point= 0.00oC -4.63oC=-4.63oC

 

See if you can now follow this video problem.

 

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