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?
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2related https://chemistry.stackexchange.com/questions/6955/deviation-from-raoults-law-because-of-adhesive-and-cohesive-forces – Mithoron Oct 21 '17 at 11:46
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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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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? – Aditya Kumar Panda Jun 05 '23 at 11:52
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1@AdityaKumarPanda such an interaction would require availability of a hydrogen bonded to an electronegative atom, which is absent in acetone. – Shivansh Jaiswal Jan 23 '24 at 12:13
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