In ethanol molecules, strong intermolecular hydrogen bonding is present because of their polar
nature. When ethanol is mixed with cyclohexane, the presence of cyclohexane molecules will
naturally reduce the hydrogen bonding and thus the escaping tendency of alcohol molecules
will increase. Same thing will happen for molecules of cyclohexane due to decreased vander
Waal’s force of attraction owing to the presence of ethanol molecules.
1.What type of deviation from Raoult’s law is exhibited by the above mentioned mixture ?
Answers
Answered by
1
Answer:
On adding cyclohexane to ethanol . Its molecules get in between the molecules of ethanol thus breaking the hydrogen bonds and reducing ethanol - ethanol interaction .
Thus will increase the vapour pressure of the solution and result in positive deviation from Rouault's law.
Answered by
0
ANSWER : delta H mix > 0
Explanation:
On adding cyclohexane to ethanol . Its molecules get in between the molecules of ethanol thus breaking the hydrogen bonds and reducing ethanol - ethanol interaction .
Thus will increase the vapour pressure of the solution and result in positive deviation from Rouault's law.
Hope it helps you ✌️ ✌️
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