grams sucrose C12H22O11 552 g water vapor pressure 2.0 mmHg less

grams sucrose C12H22O11 552 g water vapor pressure 2.0 mmHg less

 

 

How many grams of sucrose (C12H22O11) must be added to 552 g of water to give a solution with a vapor pressure 2.0 mmHg less than that of pure water at 20℃? (The vapor pressure of water at 20℃ is 17.5 mmHg.)

 

 

 

라울의 법칙

P_용액 = X_용매 × P°_용매

( 참고 https://ywpop.tistory.com/2646 )

 

 

 

P_용액 = 17.5 – 2.0

= 15.5 mmHg

 

 

 

X_용매 = P_용액 / P°_용매

= 15.5 / 17.5

= 0.8857

 

 

 

H2O의 몰질량 = 18.015 g/mol 이므로,

552 g H2O의 몰수를 계산하면,

552 g / (18.015 g/mol) = 30.64 mol (용매)

( 참고 https://ywpop.tistory.com/7738 )

 

 

 

sucrose의 몰수 = x 라 두면,

0.8857 = 30.64 / (x + 30.64)

( 참고: 몰분율 https://ywpop.tistory.com/2659 )

 

x + 30.64 = 30.64 / 0.8857

 

x = (30.64 / 0.8857) – 30.64

= 3.954 mol (용질)

 

 

 

C12H22O11의 몰질량 = 342.30 g/mol 이므로,

3.954 mol sucrose의 질량을 계산하면,

3.954 mol × (342.30 g/mol) = 1353 g

 

 

 

답: 1.35×10^3 g

 

 

 

 

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