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o-na [289]
3 years ago
12

A solution of chloroform (CHCl3) and acetone((CH3)2CO) exhibits a negative deviation from Raoult's law. This result implies that

: W. chloroform-chloroform and acetone-acetone interactions are stronger than chloroform-acetone interactions. X. chloroform-chloroform and acetone-acetone interactions are weaker than chloroform-acetone interactions. Y. acetone-acetone interactions are stronger than chloroform-chloroform interactions. Z. acetone-acetone interactions are weaker than chloroform-chloroform interactions.
Biology
2 answers:
Korolek [52]3 years ago
8 0

Answer:

The answer is W. chloroform-chloroform and acetone-acetone interactions are stronger than chloroform-acetone interactions. This is because the bond between acetone-acetone is a dipole-dipole interactions and chloroform-chloroform dipole-dipole compared to the weaker hydrogen-bonding between acetone-chloroform.

It turns out that this hydrogen-bonding happens to be stronger the original dipole-dipole forces, so this shows NEGATIVE DEVIATION from Raoult's law.

pantera1 [17]3 years ago
8 0

Answer: Option W.

Chloroform Chloroform and acetone acetone interactions are stronger than chloroform acetone interactions.

Explanation:

When chloroform and acetone mixed together, it will form a solution with negative deviation from Raoult's law because chloroform form hydrogen bond with acetone molecules. If a vapour mixture is lower that what is expected, negative deviation arises.

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