Cliffs AP Chemistry, 3rd Edition

II. H 3 BO 3

A. H 3 BO 3 is a weak acid.

B. In the equation ∆t =i ⋅m ⋅Kb, where ∆tis the boiling-point elevation, m is the mo-

lality of the solution, and Kbis the boiling-point-elevation constant for water, i (the

van’t Hoff factor) would be expected to be 4 if H 3 BO 3 were completely ionized.

According to data provided, iis about 1.5. Therefore, H 3 BO 3 must have a relatively

low Ka.

III. NaCl, MgCl 2 , and FeCl 3

A. All three compounds are chlorides known to be completely soluble in water, so

they are strong electrolytes and would increase electrical conductivities.

B. The van’t Hoff factor (i) would be expected to be 2 for NaCl, 3 for MgCl 2 , and 4

for FeCl 3.

C. Using the equation

..

mK

t

kg

mole solute

mole solute

kg

∆

0 05 0 512

100

C

B.P. of solution C

: b = :

• %

%

we find that the van’t Hoff factors for these solutions are

Compound Calculated i Expected i

NaCl 1.9 2.0

MgCl 2 2.7 3.0

FeCl 3 3.4 4.0

which are in agreement.

D. The electrical conductivity data support the rationale just provided: The greater the

number of particles, which in this case are ions, the higher the B.P.

IV. C 6 H 12 O 6

A. C 6 H 12 O 6 , glucose, is an organic molecule. It would not be expected to dissociate

into ions that would conduct electricity. The reported electrical conductivity for

glucose supports this.

B. Because C 6 H 12 O 6 does not dissociate, iis expected to be close to 1. The equation in

III. C. gives ias exactly 1.

C. The boiling-point-elevation constant of 0.512°C ⋅kg/mole would be expected to

raise the B.P. 0.0256°C for a 0.05 m solution when i=1. The data show that the

boiling-point elevation is 0.0255°C. This agrees with the theory. Therefore,

C 6 H 12 O 6 does not dissociate. With few or no ions in solution, poor electrical con-

ductivity is expected. This is supported by the evidence in the table.

Questions Commonly Asked About the AP Chemistry Exam

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