To go with Chapter 13: Silberberg Principles of General ChemistryAP Chemistry, Mrs Peck

Perform calculations with different solution concentrations such as molarity, mass percent, molality, and mole fraction.

Discuss the effects of temperature, pressure, and structure on solubility

Perform calculations with Raoult’s law Understand colligative properties such as boiling point

elevation, freezing point depression, and osmotic pressure Use colligative properties to determine the molar mass of

a soluteAP tip: Questions from this section may appear on both the F.R. and M.C. Sections of the AP exam. Question on the solubility or miscibility of substancesIn one another may appear in the essay section. Questions involving the Determination of molar mass by freezing point depression could also appear inThe lab essay.

Types of solutions A solution is a homogeneous mixture.. We will focus on liquid solutions in this topic. A solute, the substance being dissolved, is added to a solvent, which

is present in the largest amount when referring to a liquid-liquid or gas-liquid solution.

Composition Molarity: ‘M’ of a solution is the number of moles of solute per liter

of solution Mass percent: also called weight percent, is the percent by mass of

solute in a solution. Mole fraction: ‘x’ is the ratio of moles of a given component to the

total number of moles of solution. For a two component solution, where nA and nB are the moles of the two components:

xA = nA/(nA + nB) Molality: ‘m’ is the number of moles of solute per kilogram of

solvent m = moles solute/kg of solvent

A solution is prepared by mixing 30.0mL of butane (C4H10, d= 0.600 g/ml) with 65.0 mL of octane (C8H18, d= 0.700 g/ml) Assuming that the volumes add in mixing, calculate the following for butane:

A) molarity

B) mass percent

C) mole fraction

D) molality

30.0ml x 0.600g x 1 mol = 0.310 mol C4H10 = 3.26 M 1 ml 58.1g 0.0950 L

Mass solution = [30.0ml x 0.600g] + [65.0 ml x 0.700g] = 63.5g 1ml 1 mlMass solute = 30.0 ml x 0.600g = 18.0g 1mlMass % = 18 g x 100 = 28.3% butane 63.5g

Moles octane = 65.0 ml x 0.700 g x 1 mol = 0.399 mol octane 1ml 114gMole fraction butane = 0.310 mol butane = 0.437 (0.310 mol butane + 0.399 mol octane)

0.310 moles butane/0.0455 kg = 6.81m

The formation of a liquid solution begins by separating the solute into its individual components.

Next, the solvent’s intermolecular forces must be overcome to make room for the solute.

The solute and solvent then interact to form the solution.

The phrase ‘like dissolves like’ means that solubility is favored if the solute and solvent have similar polarities, as determined by their structure.

The table summarizes the solubilities of different types of solutes in different types of solvents.

Type of solute

Type of solvent

Solubility Example

Ionic Polar Usually soluble LiCl in H2O

Polar Polar Soluble (miscible)

CH3OH in H2O

Nonpolar Polar Immiscible C6H14 in H2O

Nonpolar Nonpolar Miscible C6H14 in CCl4

Discuss the solubility of each of the following solutes in carbon tetrachloride: ammonium nitrate, 1-pentanol, and pentane. Explain why each solute will or will not dissolve.

Carbon tetrachloride is nonpolar and nonpolar solutes Will dissolve in it. Ammonium nitrate is ionic and will notDissolve in a nonpolar solvent. It will dissolve in a polar solventLike water. 1-propanol is polar and will not be soluble in carbonTetrachloride. Pentane is nonpolar and will be miscible (will formA homogeneous mixture) with carbon tetrachloride.

Pressure has little effect on the solubilities of liquids and solids.

Gases become more soluble in liquids when the pressure of the gas above the liquid decreases.

The solubility of most solids increases with temperature.

However, some solids decrease solubility with temperature.

The energy needed to separate the solute-solute particles is usually greater than the energy released from the solute-solvent attractions.

It is difficult to predict the temperature dependence of the solubilities of solids,.

Gases, on the other hand, always decrease in solubility with increasing temperature.

A nonvolatile solute lowers the vapor pressure of the solvent.

The nonvolatile solute decreases the escaping tendency of the solvent molecules.

Raoult’s law: State that the vapor pressure of a solution is directly

proportional to the mole fraction of the solvent present. Psoln = xsolventP0

solvent

Psoln is the observed vapor pressure of the solution Xsolvent is the mole fraction of the solvent P0

solvent is the vapor pressure of the pure solvent. For an ideal solution, a plot of Psoln vs xsolvent at constant

temperature gives a straight line with a slope equal to P0solvent.

The vapor pressure of a solution containing 53.6g of glycerin, C3H8O3, in 133.7 g of ethanol, C2H5OH, is 113 torr at 40*C. Calculate the vapor pressure of the pure ethanol at 40*C assuming that glycerin is a nonvolatile, nonelectrolyte solute in ethanol. Xethanol = mol ethanol/ total mol in solution Pethanol = xethanol P0

ethanol

133.7 g C2H5OH x 1 mol = 2.90 mol C2H5OH 46.07 g

53.6 g C3H8O3 x 1 mol = 0.582 mol C3H8O3

92.09 g

Total mol = 0.582 mol + 2.90 mol = 3.48 mol

P0ethanol = 113 torr = 136 torr

(2.90 mol/3.48 mol)

Ideal solutions obey Raoult’s law. Real solutions best approximate the behavior of ideal

solutions when the solute concentration is low and the solute and solvent have similar types of intermolecular forces and molecular sizes.

Deviations from Raoult’s law occur when the interactions between the solute and solvent are extremely strong, as in hydrogen bonding, and the vapor pressure of the solution is lower than predicted.

Also, the vapor pressure of a real solution is greater than predicted when the intermolecular forces between the solute and solvent are weaker than the intermolecular forces between the solute-solute or solvent-solvent.

Colligative properties depend on the number, not the identity, of the solute particles in an ideal solution.

Boiling point elevation A nonvolatile solute elevated the boiling point of a solvent. Because a nonvolatile solute lowers the vapor pressure of a

solution, it must be heated to a temperature higher than the boiling point of the pure solvent.

The normal boiling point is the point what the vapor pressure of liquid equals 1 atm.

The change in boiling point due to the presence of the nonvolatile solute is represented by

ΔTb = Kbm ΔT is the difference between the boiling point of the solution

and that of the pure solvent Kb is the molal boiling point elevation constant of the solvent m is the molality of the solute in solution.

2.00 g of a large biomolecule was dissolved in 15.0g of carbon tetrachloride. The boiling point of this solution was 77.85*C. Calculate the molar mass of the biomolecule. For carbon tetrachloride, the boiling point constant is 5.03*C kg/mol, and the boiling point of pure carbon tetrachloride is 76.5*C

ΔTb = 77.85*C – 76.5*C = 1.35*C

m = 1.35*C = 0.268 mol/kg (5.03*C Kg/mol)

Mol of biomolecule = 0.0150 kg solvent x 0.268 mol hydrocarbon/kg solvent = 4.02 x 10-3 mol

Molar mass = 2.00 g = 498 g/mol 4.02 x 10-3 mol

A nonvolatile solute decreases the freezing point of a solvent.

ΔTf = Kfm ΔT is the difference between the boiling

point of the solution and that of the pure solvent.

Kf is the molal freezing point depression constant of the solvent.

Which of the following solutions would have the largest freezing point depression? Explain your answer. A) 0.010 m NaCl B) 0.010 m Na3PO4

C) 0.010 m Al2(SO4)3

D) 0.010 m C6H12O6The solution with the lowest freezing point depression is 0.010 m Al2(SO4)3, the solution which yields the highest molality (or theMost) particles when dissolved. 0.010 m Al2(SO4)3 yields five particles (ions) when dissolved or the total concentration of solute particles is0.050 m. Two particles (or 0.020 m) are present in a solution of 0.010 mNaCl. Four particles (or 0.040 m) are present in a solution of 0.010 mNa3PO4. One particle (or 0.010 m) is present in a solution of 0.010 mC6H12O6

Osmosis is the flow of a pure solvent into a solution through a semipermeable membrane.

Osmotic pressure is the pressure that must be applied to a solution to stop osmosis

π = MRT Π is the osmotic pressure in atmospheres M is the molarity of the solution R is the gas law constant, 0.08206

L*atm/mol*K T is the temperature in Kelvin

An aqueous solution of 10.00 g of catalase, an enzyme found in the liver, has a volume of 1.00 L at 27*C. The solution’s osmotic pressure at 27*C is 0.745 torr. Calculate the molar mass of the catalase.

M = π = 0.745 torr x (1 atm/760 torr) = 3.98 x 10-5 M [0.08206 x atm/mol*K] 300K

1.00 L x (3.98 x 10-5 mol/L) = 3.98 x 10-5 mol

Molar mass = 10.0 g = 2.51 x 105 g/mol 3.98 x 10-5 mol