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CH8_Page_15

# CH8_Page_15 - 0|[l839moldtg'1 —.¥ x x=ﬂ.103 mud-kg“...

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Unformatted text preview: 0|.[l839moldtg'1 —.¥ + x + x=ﬂ.103 mud-kg“ 0.0839 mol -kg" + x = 0.103 mol-kg" I = [1.619 l'nol‘ltg'I 0.019 Incl-kg" _1 x 100=23% 0.0339mo1-kg % dissociation = 8.58 Since both solutions freeze at the same temperature, we can say that 15.1} [compound C) = £in (compound D} . Given that 33'} = ivmviltf1 along with the fact that compound C has i = 1 and compound D has i a» l we can determine which compound has the larger molar mass in the following manner (for the sake of convenience we will assume that t = 2 for compound D; the overall conclusions will not change as long as we use any a") 1) : _ "c = ”D M} (c) — 1[ﬂ.lﬂﬂi;g']k‘r and Mill-3] 2[0-mkg]af 030 g 0.30 g a: u MM Md 1 t k =2 5 .t andl ‘-“ t 2 D t [0.100ng I [0.10053] 1' 0.10013 ’ owing f SD 0.30 g 0.6!] 3; ;M = ZMM .Thercfoi' dDhasth We We a c e compoun e greater molar mass. 8.60 First calculate the van’t Hoff i factor: :52} = Home 0.423 K 2i): 1.36 K-kg-mol'l x 0.124 Incl-kg" i e 1.83 The molality of all solute species (undissociated CC13COOH(uq) plus concoo-(aq) + H*(a¢1}) = 1.83 x 0.124 mol« [cg-1 .—. 0.22? mo] dig" If all the CC13COOH{an had dissociated, the total molality in solution would have been 0.248 mol ~kg" , giving an I' value equal to 2. If no 241 ...
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