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120 5 SIMPLEMIXTURES
�is expression for the volume of 1.000 kg evaluates as
V = 1000 g
0.3713 × (241.1 gmol−1) + (1 − 0.3713) × (198.2 gmol−1)
× [0.3713 × (188.2 cm3mol−1) + (1 − 0.3713) × (176.14 cm3mol−1)]
= 843.5 cm3
E5A.7(b) Consider a solution of A and B in which the fraction (by mass) of A is α.
�e total volume of a solution of A and B is calculated from the partial molar
volumes of the two components using [5A.3–144], V = nAVA + nBVB. In this
exercise V and VA are known, so the task is therefore to �nd the amount in
moles, nA and nB, of A and B in the solution of known mass density ρ.
�e mass of a volume V of the solution is ρV , so the mass of A is αρV . If
the molar mass of A is MA, then the amount in moles of A is nA = αρV/MA.
Similarly, nB = (1− α)ρV/MB.�e volume is expressed using these quantities
as
V = nAVA + nBVB =
αρVVA
MA
+ (1 − α)ρVVB
MB
�e term V cancels between the �rst and third terms to give
1 = αρVA
MA
+ (1 − α)ρVB
MB
�is equation is rearranged to give an expression for VA
VA =
MA
αρ
(1 − (1 − α)VBρ
MB
)
In this exercise letA beH2OandBbe ethanol, and as themixture is 20%ethanol
and 80% H2O by mass, α = 0.8. �e molar mass of A (H2O) is MA = 16.00 +
2 × 1.0079 = 18.0158 gmol−1 and the molar mass of B (ethanol) is MB = 2 ×
12.01 + 16.00 + 6 × 1.0079 = 46.0674 gmol−1. �e above expression for VA
evaluates as
VA =
MA
αρ
(1 − (1 − α)VBρ
MB
) = (18.0158 gmol−1)
0.8 × (0.9687 g cm−3)
×
⎛
⎝
1 − (1 − 0.8) × (52.2 cm3mol−1) × (0.9687 g cm−3)
46.0674 gmol−1
⎞
⎠
= 18.1 cm3mol−1
In evaluating this expression the density has been converted to units of g cm−3:
968.7 kg m−3 = 968.7 × 103 × 10−6 g cm−3 = 0.9687 g cm−3.