Humboldt State University ® Department of Chemistry

Richard A. Paselk

	General Chemistry	Fall 2008
Exercise: Colligative Properties of Solutions	© R. Paselk 2008
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Colligative Properties Problems - Answers

Remember that Colligative properties depend only on the number or concentration of particles in a solution. The properties are, for ideal solutions, independent of the kind or size of the particles, whether ionic, large or small etc. As a consequence:

• For ionic substances the colligative properties depend on the concentration of ions, e.g. for 1M NaCl we have 2M ions, so 2M particles.
• For all substances the colligative properties are independent of size or molecular weight, e.g. a 0.01mM solution of ribonuclease (MW = 13,700 daltons) will have the same colligative properties (bp, fp, osmotic pressure, etc.) as a 0.01mM solution of methanol (CH3OH, MW = 32).

In these exercises we will be using the equations for the various colligative properties we have studied:

• Vapor pressure lowering (Raoult's Law): P = XP°, where P° = the vapor pressure of the pure substance and X = its mole fraction.
• Boiling point elevation: T_b = K _bm, where m = molality and K_b is a proportionality constant, the molal boiling-point elevation constant specific to the solvent.
• Freezing point depression: T_f = -K_fm, where m = molality and K_f is a proportionality constant, the molal freezing-point depression constant specific to the solvent.
• Osmotic pressure (): V = nRT; or, dividing both sides by V, = MRT, where M = molarity.

Recall also the definitions of the three concentration measures fundamental to solution colligative behaviors:

• X (mole fraction) = moles solute divided by the total number of moles of solutes and solvent.
  • For example if we mixed 1.0 moles of methanol + 1.0 moles of ethanol + 3.0 moles of ethylene glycol + 5.0 moles of water the mole fraction of methanol would be, X = 1mol/(1mol + 1mol + 3mol + 5mol) = 1mol/10mol = 0.10.
  • Note that X = 1, e.g. X₁ + X₂ = X₃ = 1.
  • Note that this is the only measure we are using where ALL solution components appear explicitly.
• m (molality) = moles of solute per kilogram of solvent.
  • For example a 1.0m solution of methanol would be made by adding 1.0 mole of methanol (32.0g) to 1.000 kg of water to give 1.032 kg total.
  • If there are no other components this is easily expressed as mole fraction. Thus 1.000 x 10³g/18.02g/mol = 55.51mol water, so X = 1.0/56.51 = 0.018.
• M (molarity) = moles of solute per liter of solution.
  • For example a 1.0M solution of methanol might be made by adding 1.0 mole of methanol (32.0g) to a one L volumetric flask and then adding enough water to bring the total volume to 1.000L.

1. Find the vapor pressure of acetone (CH₃COCH₃) in a solution made by dissolving 1.00g of a non-volatile compound (sulfanilamide, C₆H₈O₂N₂S, MW = 172.1) in 10.0g of acetone. The vapor pressure of pure acetone under these conditions is 4.00 x 10³mmHg.

For a vapor pressure problem we will use the Raoult's Law equation, P = XP°

First we need to ask ourselves if the solute is covalent (moles of particles = moles) or ionic (moles of particles = moles of ions). In this case all of the atoms are hi or intermediate, so can consider covalent.

Next we need to determine the number of moles of each:

• MW_acetone = 3(12.01) + 6(1.008) + 16.00 = 58.1 g/mol; Moles_acetone = 10.00g/58.1g mol^-1 = 0.172 mol.
• Moles_{sulfanilamide} = 1.00g/172.1g mol^-1 = 5.81 x 10^-3mol.

We can now find the mole fraction of acetone, X_acetone = 0.172mol/(0.172 mol + 5.81 x 10^-3mol) = 0.967

Finally, P = XP° = (0.967)(4.00 x 10³mmHg)

P= 3.87 x 10³mmHg

 

2. Determine the total vapor pressure over a solution of 0.600 moles of toluene and 0.400 moles of benzene @ 60 °C assuming ideal behavior. The vapor pressures of pure toluene and benzene are, respectively, 139 mmHg and 392 mmHg @ 60°C.

Both of these substances are organic solvents, so covalent (the vapor pressures give away that they are not ionic since ions do not vaporize at normal temperatures).

Since the concentrations add up to 1.000 m we can use their concentrations as their mole fractions (e.g. 0.600/1.000 = 0.600)

Let's then find the vapor pressures of each solvent individually:

P_toluene = X_tolueneP°_toluene = (0.600)(139 mmHg) = 83.4 mmHg

P_benzene = X_benzeneP°_benzene = (0400)(392 mmHg) = 157 mmHg

Recalling Dalton's Law of partial pressures, P_total = P_toluene + P_benzene = 83.4 mmHg + 157 mmHg

P_total= 2.40 x 10²mmHg

 

3. Find the boiling point of a solution of 0.300 g urea (NH₂CONH₂, MW = 60.1) dissolved in 10.0 g of pure water at 1.00 atm. (K_b = 0.512 °C m^-1)

Reading the problem this appears to be a boiling point elevation problem, so T_b = K _bm

First we need to find the concentration in molality.

• The data given allow us to find moles/g and kg
  • moles_urea = 0.300g/60.1g mol^-1 = 4.99 x 10^-3 moles
  • kg_water = (10.0g)(1 kg/1,000g) = 0.0100 kg
• molality can then be found by conversion to kg: molality = (4.99 x 10^-3 moles)/(0.0100 kg) = 0.499 m

and, T_b = K _bm

T_b = (0.512 °C m^-1)(0.499 m) = 0.256°C

Finally, we can add the boiling point elevation to the boiling point of pure water at 1 atm, 100.00°C,

T = 100.256°C = 100.26°C

 

4. Calculate the molecular weight of an unknown substance if dissolving 7.39 g in 85.0 g of benzene (a non-polar solvent) raises the boiling point from 80.2 °C to 82.6 °C. (K_b = 2.52 °C m^-1)

Reading the problem this appears to be a boiling point elevation problem, so T_b = K _bm

Because benzene is a non-polar solvent we know the unknown is NOT ionic, since ionic compounds will not dissolve in non-polar solvents.

In this case we can use the equation to find the molality, which in turn we can use to find moles and thus molecular weight.

Rearranging, m = (T_b)/(K _b); m = (82.6 - 80.2 °C)/(2.52 °C m^-1) = (2.4°C)/(2.52 °C m^-1) = 0.95 m 0.95 mol/kg

And from the grams we find that (7.39g/[85.0g][1kg/1,000g]) = (7.39g)/(0.085kg) = 86.9 g/kg

Finally,

MW = (86.9 g/kg)/(0.95 mol/kg) = 91 g/mol

 

5. Calculate the boiling point of a 0.250 m aqueous solution of iron(III) chloride at 1.00 atm. (K_b = 0.512 °C m^-1)

Reading the problem this appears to be a boiling point elevation problem, so T_b = K _bm

Writing out the formula of the compound by consulting the Periodic table we get, FeCl₃ which we expect to be ionic (Fe is lo, Cl is hi).

As a result we see that we will have four moles of ions per mole of substance, so our molality, m = 4(0.250 m) = 1.00 m

The boiling point increase is then T_b = K _bm = (0.512 °C m^-1)(1.00 m) = 0.512°C.

Adding the boiling point increase to the boiling point of water at 1.00 atm we get:

 

T = 100.512°C = 100.51°C

 

6. Calculate the molecular weight of an unknown substance if dissolving 1.42 g in 25.0 g of pure benzene lowers the freezing point by 1.96°C. (K_f = 5.12 °C m^-1) [148 g/mol]

From reading the problem this is a freezing point depression problem and T_f = -K_fm

In this case we can use the equation to find the molality, which in turn we can use to find moles and thus molecular weight.

Rearranging, m = (T_f )/(-K_f) = (1.96°C)/(5.12 °C m^-1) = 0.383 m = 0.383 mol/kg solv.

Now if we find the mass/kg solv. we can find the molecular weight: 1.42g/0.025kg = 56.8g/kg solv.

Finally:

MW = (56.8g/kg solv.)/(0.383 mol/kg solv.) = 148 g mol^-1

7. What is the freezing point of a 0.100 m aqueous solution of aluminum chloride? (K_f = 1.988 °C m^-1)

Reading the problem this appears to be a freezing point depression problem and T_f = -K_fm

Writing out the formula of the compound by consulting the Periodic table we get, AlCl₃ which we expect to be ionic (Al is lo, Cl is hi).

As a result we see that we will have four moles of ions per mole of substance, so our molality, m = 4(0.100 m) = 0.400 m

The freezing point depression is then: T_f = -K_fm = (-1.988°C m^-1)(0.400 m) = -0.7952 °C

So the freezing point becomes 0.000 -7.952 °C

T_f = -0.795°C

 

8. A 0.150 m solution of acetic acid in water is found to have a freezing point of - 0.28 °C. What is the concentration of hydrogen ions in this weak acid solution? (K_f = 1.855 °C m^-1)

For this problem we want to recall that acetic acid is a weak acid, so the molality will be the sum of the acid plus the hydrogen ion and acetate ion concentrations.

From the data we can find the total molality. Rearranging the freezing point depression equation, T_f = -K_fm, we get: m = (T_f )/(-K_f)

Solving with the data provided, m = (-0.28°C)/(1.855 °C m^-1) = 0.1509 m

In this case we started by adding 0.150 moles of acetic acid, which then partially dissociated. So after dissociation, assuming x is the amount dissociated we have:

• acetic acid = 0.150 - x
• acetate ion = x
• hydrogen ion = x
• total moles then = 0.150 - x + 2x = 0.150 + x

Thus hydrogen ion concentration, [H⁺] = 0.1509 m- 0.150 m

[H⁺] = 0.009 m

 

9. Find the osmotic pressure of a 0.500 M solution of sodium chloride at 0 °C.

= MRT

Of course sodiium chloride is ionic so the molarity of ions is double the stated molarity, 2(0.500 M) = 1.000 M = 1.000 mol L^-1

Plugging values into the equation we get = (1.000 mol L^-1)(0.0821 L atm mol^-1 K^-1)(273.15 K)

= 22.4 atm

 

10. In order to find the molecular weight of hemoglobin 0.500 g was dissolved in enough water in a volumetric flask to give 100.0 mL of solution. The osmotic pressure of this solution was then measured at 25°C and found to be 1.35 mmHg. Calculate the moleculaar weight.

= MRT

To find the molecular weight we need to have moles and grams. We are given grams in the problem, so first let's find the moles via finding M

Rearranging the osmotic pressure equation we get M = /RT. Substituting values then we get:

M = {(1.35 mmHg)/(760 mmHg atm^-1)}/{(0.0821 L atm mol^-1 K^-1)(273.15 +25)K = (1.78 x 10^-3 atm)/{(0.0821 L atm mol^-1 K^-1)(278 K)}

M = 7.275 x 10^-5 mol/L

From the problem we see that the solution had 0.500 g/100.0 mL = 5.00g/L

To get g/mol we then can use dimensional analysis to set up the equation:

MW = (5.00 g/L)/(7.275 x 10^-5 mol/L) = 6.87 x 10⁴ g/mol

In addition to these exercises you should familiarize yourself with the text materials.

© R A Paselk

Last modified 6 December 2008