ethanol 60℃ 0.459 atm 10.0 mmol naphthalene 90.0 mmol ethanol

ethanol 60℃ 0.459 atm 10.0 mmol naphthalene 90.0 mmol ethanol

 

 

The vapor pressure of pure ethanol at 60℃ is 0.459 atm.

Raoult’s Law predicts that a solution prepared by

dissolving 10.0 mmol naphthalene (nonvolatile) in 90.0 mmol ethanol

will have a vapor pressure of _____ atm.

 

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에탄올의 몰분율, X_용매

= 90.0 / (10.0 + 90.0)

= 0.9

 

 

 

P_용액 = X_용매 × P°_용매

( 참고 https://ywpop.tistory.com/2646 )

 

= 0.9 × 0.459 atm

= 0.413 atm

 

 

 

답: 0.413 atm