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Section 9.2
Partial Molar Quantities

Figure 9.3
Volumes at 20°C and 1 atm of
solutions containing 1000 g of
water and n moles of MgSO . The
4
dashed lines are used to find that

3
V 1.0 cm /mol at molality
MgSO 4
0.1 mol/kg.

This key result expresses the solution’s volume V in terms of the partial molar volumes

V i of the components of the solution, where each V i [Eq. (9.9)] is evaluated at the tem-
perature, pressure, and mole fractions of the solution.

Equation (9.16) is sometimes written as V © x V , where the mean molar
m i i i
volume V of the solution is [Eq. (8.11)] V V/n, with n © i n .
m
i
m
The change in volume on mixing the solution from its pure components at con-
stant T and P is given by the difference of (9.16) and (9.4):

¢ V V V* a n 1V V* 2 const. T, P (9.17)
i
m,i
i
mix
i
where mix stands for mixing (and not for mixture).
Measurement of Partial Molar Volumes
Consider a solution composed of substances A and B. To measure V B
10V>0n 2 , we prepare solutions at the desired T and P all of which contain a fixed
B T,P,n A
number of moles of component A but varying values of n . We then plot the measured
B
solution volumes V versus n . The slope of the V-versus-n curve at any composition
B
B

is then V B for that composition. The slope at any point on a curve is found by drawing
the tangent line at that point and measuring its slope.

Once V B has been found by the slope method, V A can be calculated from V and

B
A
V B using V n V A n V B [Eq. (9.16)].
Figure 9.3 plots V versus n(MgSO ) for MgSO (aq) solutions that contain a fixed
4 4
amount (1000 g or 55.5 mol) of the solvent (H O) at 20°C and 1 atm. For 1000 g of
2
solvent, n is numerically equal to the solute molality in mol/kg.
B
EXAMPLE 9.3 The slope method for partial molar volume

Use Fig. 9.3 to find V and V in MgSO (aq) at 20°C and 1 atm with
MgSO 4 H 2 O 4
molality 0.1 mol/kg.
Drawing the tangent line at 0.1 mol of MgSO per kg of H O, one finds its
4 2
3
3
slope to be 1.0 cm /mol, as shown in Fig. 9.3. Hence V 1.0 cm /mol at
MgSO 4
m 0.1 mol/kg. At m 0.1 mol/kg, the solution’s volume is V
MgSO 4 MgSO 4
3
1001.70 cm . This solution has 0.10 mol of MgSO and 1000 g of H O, which

2
4
is 55.51 mol of H O. Use of V n V n V gives
2 A A B B
3
3
V 1001.70 cm 155.51 mol2V H 2 O 10.10 mol211.0 cm >mol2
3
V H 2 O 18.04 cm >mol