Lecture_5b

5 ml mol 1 therefore using eq 533 we get v n e v e

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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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This document was uploaded on 02/28/2014 for the course CHEM 311 at LA Tech.

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