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**Unformatted text preview: **x E = 0.207; x W = 0.793.
From the figure, we estimate that at
these mole fractions, the partial molar
volumes are
V E = 55.0 mL mol −1 ;
V W = 17.5 mL mol −1 .
Therefore, using Eq. (5.33), we get
V = n E V E + n W V W = 8.68 55.0 + 33.30 17.5
= 1060 mL.
If the solution was ideal, we would use the molar volumes of the pure substances
to obtain
46.07 g mol −1
18.02 g mol −1
V = 8.68 mol
+ 33.30 mol
0.785 g mL −1
0.997 g mL −1
= 1111 mL.
Therefore, we see that the non-ideal nature of the solution is reflected in a
contraction of volume by 51 mL. Another example of applying Eq. (5.33):
Density of a 50% by mass solution of ethanol in water at 25°C is 0.914 g mL–1.
Given that the partial molar volume of water at this composition is 17.4 mL
mol–1,what is the partial molar volume of ethanol?
No. of moles of ethanol in 100 g of solution: 50 g/46.07 g mol–1 = 1.085 mol.
No. of moles of water in 100 g of solution: 50 g/18.02 g mol–1 = 2.775 mol.
Now, since V = n E V E + n W V W ,...

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