2.3 Analysis
2.3.1 Peak Fitting
You should recall we discussed pure and impure substances even when they are homogenous or heterogenous systems. If a pure liquid is placed in a closed container that is initially evacuated. the liquid evaporates to fill up the space above the liquid. At any given temperature when equilibrium is established. the pressure exerted by the vapours of a liquid is called the vapour pressure of the pure liquid (P", ). If a non-volatile solute is added to this liquid. the equilibrium vapour pressure (p,) over the solution is found to be less than that of the pure liquid. According to Raoult's law. the vapour pressure (p 1 ) of the solvent of an ideal solution is given by
P, Po, x, ...3.1
Where x, is the mole fraction of the solvent in the solution and p", is the vapour pressure of the pure solute. Since the solute is non-volatile. its contribution towards the vapour pressure of the solution is negligible. Hence. the vapour pressure of solution containing a non-volatile solute is only due to the solvent. Since the mole fraction (x 1 ) of the solvent is less than one, it is clear ti'om Eq. 3.1 that p will 1 also be less than p0,. The lowering of vapour pressure when a non-volatile solute is added to a solvent is given by (p0 - ).
1
Substitution for from Eq. 3.1. we get
Po,- P, Po,- Po, x,= Po, ( I - x,) ..., ... 3.2 Since 1 + x,I.
3.3 Where x2 is the mole fraction of the solute.
According to Eq.3.3, the lowering of vapour pressure of the solvent depends both on the vapour pressure of the pure solvent and on the mole fraction of the solute in the solution. In other words. 11
+
1 2
1
0
depends on the nature of the solvent and on the concentration of the solute and not on the nature of the solute. Eq.3.3 can be written as
Po,- P, ... 3.4 -.-- =x2
Po,
The expression on the left hand side of Eq.3.4 is called the relative lowering of vapour pressure.
Eq.3.4 also may be stated as:
"The relative lowering of vapour pressure of a solution is equal to the mole fraction of the solute''.
'.fhe above statement is also called Raoult's law of relative lowering of vapour pressure.
3.f.l Determination of Relative Molecular Mass from the Lowering of Vapour Pressure Having discussed relative loweringofvapour pressure, it is possible to measure the relative molecular mass of a non-volatile solute, if we know the vapour pressure of a dilute solution containing.the non·
volatile solute. Now let us consider the equation that we need to measure the relative mole£.ular mass from the lowering of vapour pressure.
Assuming w kg of the solute having molar mass M, kg mol·' is dissolved in w kg pf the solvent having molar mass M kg mol·', then from Eq.3.4 we have:
Po, -·p, Po,
=x=-
n,
2 n n
I 2
...3.5
p 1 - p1 w,fM2
or =-
- p0 w, 1M,
+
w,fM2 ... 3.61
For a dilute solution, the number of moles of the solute (n,) is negligible in comparison to the number of moles of solvent (n ). Thus Eq.3.6 reduces to ·
Po,-p, w,fM, Po, w,IM,
w, M, ...3.7 w, M,
Thus; if we are able to know the mass of the solvent, its relative molecular mass, mass of the solute.
the vapour pressure of the solution and of the solvent, the relative molecular mass of the solute can be calculated from Eq.3.7.
3.1.2 easurement of Vapour Pressure Lowering
The lowering of vapour pressure 'of a solvent when a non-volatile solute is dissolved in it can be measured by the following methods:
(i) Static Method
In this method, the difference between the vapour pressure of the solvent and the solution is determined with the help of a differential manometer. The apparatus is as shown in Fig.6.1. It consists of two bulbs which are connected to a manometer. One arm of the manometer is connected with the bulb, A, containing the solvent and the other arm with the bulb, B, coniainingthe solution.
The manometric liquid is an inert, non-volatile, low density liquid such as b-bromonaphthalene.
-0
From the difference in the levels of the liquid in the two arms, the difference in vapour pressure between the solvent and the solution can be read directly.
Fig. 6.1: Static Methodfin· determining the lowering of vapour pressure
(ii) Dynamic Method
This method is due to Ostwald and Walker. In this method (Fig.6.2), a stream of dry and purified air is first passed through a pre-weighed set of bulbs containing the solution. The air coming out of the bulbs takes up an amount of vapour which is proportional to the vapour pressure of the solution.
Evidently, the loss in mass suffered by tile bulbs containing the solution is proportional to the vapour pressure of the solution. lfw, is the loss in mass suffered by the solution bulbs and p, is the vapour pressure of the solution. then we have.
w-\ (l p, 3.8
\
IV
Fig. 6.2: Ostwald and Walker\apparatus for measuring the vapour pressure lowering:I11 air: 1!11 solution hulhs: Ill/; soh·enl hulhs: r!VJ H"eighed CaC IUhes. rV; cily cw·
The air coming out ofthe solution bulbs is then passed through pre-weighed set of bulbs containing the solvent. As the air containing vapours of the solvent passes through the solvent. it takes up a further amount of vapour which is proportional to the difference in vapour pressures of the solvent and the solution. Evidently the loss in mass suffered by the solvent bulbs is proportional to the difference in vapour pressures oft he pure solvent and the solution. If w 8 is the loss in mass suffered by the solvent- bulbs and p", is the vapour pressure of the solvent, then we have
WB
a
Po,- P, ...3.9)
)
From Eqs.3.8 and 3.9 we have (w, + w 8
a
p, + p0 ,- P,or(w,+w )a p", ... 3.10
8
From Eqs.3.8 and 3.10 we have.
3.11
Thus knowing w,and w 0 , the relative Jo,wering of vapour pressure can be calculated using Eq.3.11.
lfthe solvent is water, the moist air coming out of the solvent bulbs may be passed through a weighed tube containing fused calcium chloride which absorbs moisture from moist air. The mass of calcium chloride tube will increase and the increase in mass will be equal to (w,+ w 8 .
Example I
The vapour pressure of a solution containing 0.0 12 kg of a solute dissolved in 0. I 00 kg of water at 300 K is 3.50 X I O' Pa. Calculate the molar mass of the solute, if the vapour pressure of water at 300 K is 3.70 x 103 Pa.
Solution w, O.IOOkg M,0.018 kg mot·'
w, O.OI2 kg M
' .
p0 3.70 X I 0' Pa
1 P, 3.50 x 10 3 Pa
Using Eq.3.7. we have
(3.70 x I 0' Pa- 3.50 x I O' Pa) = (0.0 12 kg)(O.OI8 kg mot·')
3.70x IO'Pa O.IOOkg(M,)
or M (0.012 kg) (0.018 kg mol·') x 3.70 X 10' Pa
' O.IOOkg 0.20 x 10' Pa
0.040.kg mol·'
In this section, we discussed the correlation between the relative lowering of vapour pressure and the mole fraction of a non-volatile solute in a solution. In the next section, we shall study how the boiling.
point of a solvent is suffered when a non-volatile solute is added toot.
Exercise I
State whether the following statements are true or false:
(i) Lowering of vapour pressure is not a colligative property.
(ii) Relative lowering of vapour pressure is equal to the mole fraction of the solvent.
(iii) Relative lowering of vapour pressure de(>ends upon the number of particles of the solute and the solvent.
0