25℃ benzene vapor 0.1252 atm 6.50 g C10H8 76.0 g benzene

25℃ benzene vapor 0.1252 atm 6.50 g C10H8 76.0 g benzene

 

 

At 25℃, the vapor pressure of pure benzene is 0.1252 atm.

When 6.50 g of a solid sample of naphthalene, C10H8 (128.17 g/mol),

is dissolved in 76.0 g of pure benzene, C6H6 (78.0 g/mol),

calculate the vapor pressure of benzene above the solution.

 

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각 물질의 몰수를 계산하면,

6.50 g / (128.17 g/mol) = 0.0507 mol C10H8

( 참고 https://ywpop.tistory.com/7738 )

 

76.0 / 78.0 = 0.9744 mol C6H6

 

 

 

용매인 벤젠의 몰분율을 계산하면,

0.9744 / (0.9744 + 0.0507) = 0.9505

( 참고 https://ywpop.tistory.com/2659 )

 

 

 

라울의 법칙

P_용매 = X_용매 × P°_용매

( 참고 https://ywpop.tistory.com/2646 )

 

= 0.9505 × 0.1252 atm

= 0.119 atm

 

 

 

답: 0.119 atm

 

 

 

 

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