Chloroform is a polar molecule, and benzene is nonpolar. Shouldn't the chloroform-chloroform and benzene-benzene intermolecular forces be stronger than chloroform-benzene interactions (like dissolves like), which would result in a positive deviation from Raoult's law?
1 Answer
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This is because when these liquids are mixed, H-bonding type interactions are formed between hydrogen atom of chloroform (partial positive charge due to 3 Cl atoms) and 'pi' electron cloud of benzene ring. Thus, chloroform-benzene interactions are stronger than chloroform-chloroform and benzene-benzene interactions.
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$\begingroup$ Then, why does acetone and benzene show positive deviation? Doesn't acetone (being a polar molecule) form hydrogen bonds with the "pi" electron cloud of the benzene ring? $\endgroup$ Commented Jun 5, 2023 at 11:52
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1$\begingroup$ @AdityaKumarPanda such an interaction would require availability of a hydrogen bonded to an electronegative atom, which is absent in acetone. $\endgroup$ Commented Jan 23 at 12:13
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$\begingroup$ @ShivanshJaiswal Yeah, I've got it. Thanks! $\endgroup$ Commented Jan 28 at 10:52