Boiling Point Elevation Problems


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Example #1: What is the boiling point elevation when 11.4 g of ammonia (NH3) is dissolved in 200. g of water? Kb for water is 0.52 °C/m.

Solution

1) Determine molality of 11.4 g of ammonia in 200. g of water:

11.4 g / 17.031 g/mol = 0.6693676 mol

0.6693676 mol / 0.200 kg = 3.3468 m

2) Determine bp elevation:

Δt = i Kb m

Δt = (1) (0.52 °C/m) (3.3468 m)

Δt = 1.74 °C


Example #2: 0.64 g of adrenaline in 36.0 g of CCl4 produces a bp elevation of 0.49 °C. What is adrenaline's molecular weight?

Solution

1) determine number of moles of adrenaline in solution:

ΔT = i Kb m

0.49 °C = (1) (4.95 °C / m) (x / 0.0360 kg)

x = 0.0035636 mol

From here it is a simple step: grams divided by moles equals the desired answer.

0.64 g / 0.0035636 mol = 180 g/mol (to two sig figs)

Notice that the Kb value for CCl4 was not given in the problem. In a textbook, the value would be in a chapter table or in an appendix. On ye olde Intertubes, one must unleash the Great Googlizer.

By the way, the textbook value for adrenaline's molecular weight is 183.204 g/mol


Example #3: Seawater is about 3.5% (by weight) dissolved solids, almost all of which is NaCl. Calculate the normal boiling point of seawater.

Solution

Some comments prior to starting:

a) 3.5% means 3.5 grams of solids per 100 grams total of solution. So that means 3.5 grams of solids are dissolved in 96.5 grams of water.
c) since we know most of the solids are NaCl, we can't go too wrong is we assume NaCl is 100%. Some error is introduced, but not too much.

1) Calculate moles of NaCl:

3.5 g / 58.5 g mol¯1 = 0.0598 mol

2) Calculate molality of NaCl:

m = 0.0598 mol / 0.0965 kg = 0.612 m

3) use boiling point elevation constant:

ΔT = i Kb m

x = (2) (0.52 °C m¯1 ) (0.612 m) = 0.64 °C

So, the water boils at 100.64 °C


Example #4: 2.60 grams of a compound know to contain only indium and chlorine is dissolved in 50.0 g of tin(IV) chloride (Kb = 9.43 °C kg mol¯1). The normal boilng point is raised from 114.1 °C for pure SnCl4 to 116.3 °C for the solution. What is the molecular weight and probable molecular formula for the solute?

Solution

1) Keep in mind that the units on molecular weight are grams per mole. We already have the grams, what we now need are the moles that 2.60 grams represent. We can get that by the molality (moles solute / kg solvent):

ΔT = i Kb m

2.2 = 9.42 (x / 0.050); x = 0.01168 mol

2) Now we calculate the molecular weight:

2.60 g / 0.01168 mol = 222.6 g/mol

The probable molecular formula is InCl3, but I will let you ponder on how I did it.


Example #5: 5.00 g of an organic solid is dissolved in 100.0 g of benzene. The boiling temperature of this solution is 82.42 °C. The organic compound is 15.72% nitrogen, 7.92% hydrogen, 35.92% oxygen and the remainder is carbon. The boiling temperature of pure benzene is 80.1 °C; Kb = 2.53 °C-kg/mol.

a. Determine the molecular weight of the organic solid.
b. Determine the molecular formula of the solid
c. Determine the mole fraction of the organic solid in the solution
d. If the density of this solution is 0.8989 g/mL , calculate the molarity of the solution

Solution:

1) Determine moles of solute using boiling point elevation data:

Δt = Kb m

1.42 °C = 2.53 °C-kg/mol (x / 0.100 kg)

1.42 °C = (25.3 °C/mol) (x)

x = 0.0561265 mol

2) Use moles and 5.00 g to determine molecular weight (answer to part a):

5.00 g / 0.0561265 mol = 89.0845 g/mol

89.1 g/mol (to 3 sig figs)

3) Use an on-line empirical formula calculator to determine the empirical formula:

C3H7NO2

4) Determine the molecular formula (answer to part b):

the "empirical formula weight" of C3H7NO2 is 89.1 g

the molecular weight divided by the "EFW" yields 1

therefore, the molecular formula is:

C3H7NO2
the same as the empirical formula

5) Determine the mole fraction of the solute (answer to part c):

moles of solvent = 100.0 g / 78.1134 g/mol = 1.28019 mol

χsolute = 0.0561265 mol / (0.0561265 mol + 1.28019 mol)

χsolute = 0.042

6) Determine molarity of solution (answer to part d):

i) our solution weighs 105.0 g; calculate its volume:
0.8989 g/mL = 105.0 g / x

x = 116.81 mL

ii) calculate the molarity:
0.0561265 mol / 0.11681 L = 0.48 M

Example #6: A 5.00 g sample of a large biomolecule was dissolved in 16.0 g of carbon tetrachloride. The boiling point of this solution was determined to be 77.85C. Calculate the molar mass of the biomolecule. For carbon tetrachloride, the boiling-point constant is 5.03Ckg/mol, and the boiling point of pure carbon tetrachloride is 76.50C.

Example #7: An organic compound contains: C, 18.3%; H, 0.51%; Br, 81.2%. When 0.793 g of the compound was dissolved in 14.80 mL of chloroform (density = 1.485 g/mL), the solution boiled at 60.63 °C. Pure chloroform boils at 60.30 °C and has Kb = 3.63 °C kg/mol. What is the molecular formula of the organic compound?

Solution:


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