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**Unformatted text preview: **es is a 1 molal solution, i.e., a
solution that contains 1 mol of the substance in 1.0 kg of solvent.
7 Physical Interpretation of partial molar quantities:
It may appear that there is something “not quite right” about the following two
equations:
V = n 1 V 1 + n 2 V 2 , where
, and V 2 = æ ¹V ö
.
V 1 = æ ¹V ö
ç ¹n ÷
ç ¹n ÷
è 1 ø P, T , n 2
è 2 ø P, T , n 1
Based on what we have seen so far, the first equation should be
dV = V 1 dn 1 + V 2 dn 2 ,which is simply another application of the chain rule in partial
differentiation.
However, Eq. (5.33) can indeed be justified on physical grounds as follows.
Consider a large volume of solution containing ethanol (E) and water (W). We
now add a small amount of water, say, ∆nW moles of water, to this solution. We
would want to express the new volume of the solution as
V new = V old + Dn W V ,m ,
W
where V ,m is the molar volume of pure water. However, this will give us the final
W
volume only in the case of an ideal solution. In the ethanol-water solution, the
effective molar volumes of both substances are different from their molar volumes
in the absence of the other substance. Designating...

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