Custom Search

 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.

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

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

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 pointi= van't Hoff factor m= molality (moles solute/ kg solvent) Kb= boiling point elevation constant (water Kb=+0.52oC/m) DTf= change in freezing pointi= 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

Example 1- van't Hoff   i= 1 (nonelectrolytes)

A solution is made by dissolving 10.20 grams of glucose (C6H12O6) in 355 grams of water. What is the freezing-point depression of the solvent if the freezing point constant is -1.86 °C/m?

DTf=imKf

i=1 (see table 2);    m= mol/kg

moles=10.20g (1mol/180g)=0.0556mol                kg=355g(1 kg/1000g)=0.355kg

m= mol/kg= 0.0556 mol/0.355kg= 0.156mol/kg;

Kf=-1.86oC/m (see table 1)

DTf=1(0.156mol/kg)(-1.86oC/m)= -0.291oC

Table 2-Van't Hoff Factors (i)

 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

Example 2- van't Hoff   i= 3 (electrolytes)

Determine the freezing point of a solution made from 3.0 kgs of water and 2.5mol of CaCl2.

DTf=imKf

i=3 (see table 2);     m= mol/kg= 2.5 mol/3.0 kg= 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

Example 3- Boiling Point elevation; van't Hoff  i=2 (electrolyte)

Determine the boiling point of a solution made from 2,000.g of water and 58.5g mol of NaCl.

DTb=imKb

i=2 (see table 2);

m= mol/kg

mol NaCL= 58.5g(1mol/58.5g)=1.00mol NaCl    kg=2,000.g(1 kg/1000g)=2.000kg

m= mol/kg= 1.00 mol/2.000kg= 0.5mol/kg;       Kb=+0.52oC/m (see table 1)

DTb= (2)(0.5mol/kg)(+0.52oC/m) =+0.52oC

Boiling point= 100.oC +0.52oC=100.52oC

back to Math of Chemistry Links