The thermodynamic description of mixtures
3.7 Measures of concentration
Interspersed among the phospholipids of biological membranes are sterols, such as cholesterol (8), which is largely hydrophobic but does contain a hydrophilic –OH group. Sterols, which are present in different proportions in different types of cells, prevent the hydrophobic chains of lipids from “freezing” into a gel and, by dis- rupting the packing of the chains, spread the melting point of the membrane over a range of temperatures.
SELF-TEST 3.4 Organisms are capable of biosynthesizing lipids of different composition so that cell membranes have melting temperatures close to the am- bient temperature. Why do bacterial and plant cells grown at low temperatures synthesize more phospholipids with unsaturated chains than do cells grown at higher temperatures?
Answer:Insertion of lipids with unsaturated chains lowers the plasma membrane’s melting temperature to a value that is close to the lower ambient temperature.
in medicine. A useful measure of concentration of a gas J in a mixture is its mole fraction, the amount of J molecules expressed as a fraction of the total amount of molecules in the mixture. In a mixture that consists of nAA molecules, nBB mol- ecules, and so on (where the nJare amounts in moles), the mole fraction of J (where JA, B, . . .) is
Amount of J (mol)
xJ
n n
J (3.4a)
Total amount of molecules (mol)
wherennAnB . For a binary mixture, one that consists of two species, this general expression becomes
xA nA
n
AnB xB nA
n
BnB xAxB1 (3.4b)
When only A is present, xA1 and xB0. When only B is present, xB1 and xA0. When both are present in the same amounts, xA1⁄2 and xB1⁄2. (Fig. 3.19).
SELF-TEST 3.5 Calculate the mole fractions of N2, O2, and Ar in dry air at sea level, given that 100.0 g of air consists of 75.5 g of N2, 23.2 g of O2, and 1.3 g of Ar. (Hint:Begin by converting each mass to an amount in moles.) Answer:0.780, 0.210, 0.009
We need to be able to assess the contribution that each component of a gaseous mixture makes to the total pressure. In the early nineteenth century, John Dalton carried out a series of experiments that led him to formulate what has become known as Dalton’s law:
The pressure exerted by a mixture of perfect gases is the sum of the pressures that each gas would exert if it were alone in the container at the same temperature:
ppApB (3.5)
xB= 0.167 A B xA= 0.167 xB= 0.833
xA= 0.452 xB= 0.548
xA= 0.833
Fig. 3.19 A representation of the meaning of mole fraction. In each case, a small square represents one molecule of A (gray squares) or B (white squares). There are 84 squares in each sample.
In this expression, pJis the pressure that the gas J would exert if it were alone in the container at the same temperature. Dalton’s law is strictly valid only for mix- tures of perfect gases, but it can be treated as valid under most conditions we encounter.
For any type of gas (perfect or not) in a mixture, the partial pressure,pJ, of the gas J is defined as
pJxJp (3.6)
wherexJis the mole fraction of the gas J in the mixture. For perfect gases, the par- tial pressure of a gas defined in this way is also the pressure that the gas would ex- ert if it were alone in the container at the same temperature. Moreover, defined in this way, eqn 3.5 is true for mixtures of real gases as well as perfect gases, but the partial pressure so defined is no longer the pressure that a gas would exert if it were alone in the container.
ILLUSTRATION 3.1 Calculating partial pressures of the gases in air FromSelf-test3.5, we have xN20.780,xO20.210, and xAr0.009 for dry air at sea level. It then follows from eqn 3.6 that when the total atmospheric pres- sure is 100 kPa, the partial pressure of nitrogen is
pN2xN2p0.780 (100 kPa)78.0 kPa
Similarly, for the other two components we find pO221.0 kPa and pAr 0.9 kPa. ■
Three measures of concentration are commonly used to describe the composi- tion of mixtures of liquids or of solids dissolved in liquids. One, the molar concen- tration, is used when we need to know the amount of solute in a sample of known volume of solution. The other two, the mole fraction, which we already encoun- tered (eqn 3.4), and the molality, are used when we need to know the relative num- bers of solute and solvent molecules in a sample.
Themolar concentration, [J] or cJ, of a solute J in a solution (more formally, the “amount of substance concentration”) is the chemical amount of J divided by the volume of the solution:4
Amount of J (mol)
[J] n V
J (3.7)
Volume of solution (L)
Molar concentration is typically reported in moles per liter (mol L1; more for- mally, as mol dm3). The unit 1 mol L1is commonly denoted 1 M(and read “mo- lar”). Once we know the molar concentration of a solute, we can calculate the amount of that substance in a given volume, V, of solution by writing
nJ[ J]V (3.8)
4Molar concentration is still widely called “molarity.”
Themolality,bJ, of a solute J in a solution is the amount of substance divided by the mass of solvent used to prepare the solution:
Amount of J (mol)
bJ
mso n
lv J
ent
(3.9)
Mass of solvent (kg)
Molality is typically reported in moles of solute per kilogram of solvent (mol kg1).
This unit is sometimes (but unofficially) denoted m, with 1 m1 mol kg1. An important distinction between molar concentration and molality is that whereas the former is defined in terms of the volume of the solution, the molality is defined in terms of the mass of solvent used to prepare the solution. A distinction to re- member is that molar concentration varies with temperature as the solution ex- pands and contracts, but the molality does not. For dilute solutions in water, the numerical values of the molarity and molar concentration differ very little because 1 liter of solution is mostly water and has a mass close to 1 kg; for concentrated aqueous solutions and for all nonaqueous solutions with densities different from 1 g mL1, the two values are very different.
As we have indicated, we use molality when we need to emphasize the rela- tive amounts of solute and solvent molecules. To see why this is so, we note that the mass of solvent is proportional to the amount of solvent molecules present, so from eqn 3.9 we see that the molality is proportional to the ratio of the amounts of solute and solvent molecules. For example, any 1.0 maqueous nonelectrolyte so- lution contains 1.0 mol solute particles per 55.5 mol H2O molecules, so in each case there is 1 solute molecule per 55.5 solvent molecules.
EXAMPLE 3.2 Relating mole fraction and molality
What is the mole fraction of glycine molecules in 0.140 mNH2CH2COOH(aq)?
Disregard the effects of protonation and deprotonation.
Strategy We consider a sample that contains (exactly) 1 kg of solvent and hence an amount nJbJ (1 kg) of solute molecules. The amount of solvent molecules in exactly 1 kg of solvent is
nsolvent 1 M
kg
whereMis the molar mass of the solvent. Once these two amounts are available, we can calculate the mole fraction by using eqn 3.4 with nnJnsolvent. Solution It follows from the discussion in the Strategy that the amount of glycine (gly) molecules in exactly 1 kg of solvent is
ngly(0.140 mol kg1) (1 kg)0.140 mol
The amount of water molecules in exactly 1 kg (103g) of water is
nwater 103 mol
18.02 103g
18.02 g mol1
The total amount of molecules present is
n0.140 mol mol
The mole fraction of glycine molecules is therefore
xgly 2.52 103
A note on good practice:We refer to exactly1 kg of solvent to avoid problems with significant figures.
SELF-TEST 3.6 Calculate the mole fraction of sucrose molecules in 1.22 m C12H22O11(aq).
Answer:2.15 102■