Page 138 - Modern physical chemistry

6.13 Elevation of the Boiling Point 129

number of equivalents of the constituent in one liter of the phase. The unit equivalent
per liter is abbreviated as N.
When a phase is kept at constant pressure, the densities of constituents vary with
temperature. To circumvent this effect, a person bases concentrations on a mass or
number of moles present. Thus, the molal concentration or molality (mJ of a solute is
the number of moles of the solute in 1000 g solvent. The unit mole per 1000 g solvent is
abbreviated as m.
The mole fraction (Xi) for a constituent is the fraction of the total moles that is moles
of the constituent, in the given region. This also equals the number fraction of the mol-
ecules that are the constituent molecules. The mass fraction equals the mass of the con-
stituent divided by the total mass in the chosen region. This may be multiplied by 100 to
give a percentage.
To construct Xi from m i one calculates the total number of moles in the solution con-
taining m i moles of substance i, then divides m i by this value. To construct Xi from C i ,
one calculates the total number of moles in one lifer of solution, then divides C i by this
number. Thus, we get

X. = mi = Ci [6.65]
t 10001 Ms +mi (lOOOd-CiMi)1 Ms +Ci '

where Ms is the molecular mass of the solvent, Mi the molecular mass of solute i, d the
mass density of the solution, Xi the mole fraction of solute i, m i the molality of i, and c i
the molarity of i.

6. 13 Elevation of the Boiling Point
Adding a nonvolatile solute to a given solvent lowers the vapor pressure at each tem-
perature. Consequently, the whole vapor pressure curve is lowered. Furthermore, the
temperature at which the vapor reaches a given pressure is raised. See figure 6.20.
Consider a solvent whose vapor pressure is PJ. This varies with temperature follow-
ing the Clausius - Clapeyron equation

dP =~dT, [6.66]
O
pO RT2
in which Lv is the heat of vaporization while T is the temperature and R the gas constant.
Let a nonvolatile solute be added while the vapor pressure is kept constant. From
formula (6.59). we have
[6.67]
Differentiating (6.67) yields
[6.68]

or
dP o
dXA
--=--- [6.69]
X A pO
Eliminating dPJIP from (6.66) and (6.69), then rearranging, leads to

RT2
dT=--- dX A • [6.70]
LvXA