fructose 1.049 g/cm3 osmotic 17.0 atm 25℃ freeze

fructose 1.049 g/cm3 osmotic 17.0 atm 25℃ freeze

 

 

An aqueous fructose solution having a density of 1.049 g/cm3

is found to have an osmotic pressure of 17.0 atm at 25℃.

Find the temperature at which this solution freezes.

Given: for water Kf = 1.86 ℃/m;

molecular mass of fructose = 180.16 g/mol

A) –1.30℃

B) –1.41℃

C) –1.52℃

D) –1.57℃

E) –1.69℃

 

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삼투압, π = MRT

( 참고 https://ywpop.tistory.com/1921 )

 

M = π / RT

= (17.0) / [(0.08206) (273.15+25)]

= 0.6948 M

 

 

 

0.6948 M 용액 1 L 속 용질의 몰수

= (0.6948 mol/L) (1 L) = 0.6948 mol fructose (용질)

 

 

 

0.6948 mol × (180.16 g/mol) = 125.175 g fructose

---> 용질의 질량

 

 

 

용액 1 L의 질량

= 1000 mL × (1.049 g/mL) = 1049 g 용액

 

 

 

용매의 질량

= 1049 – 125.175 = 923.825 g 용매

 

 

 

몰랄농도 = 용질 mol수 / 용매 kg수

= 0.6948 mol / (923.825/1000) kg

= 0.752 m

 

 

 

어는점 내림, ΔTf = Kf × m

( 참고 https://ywpop.tistory.com/1920 )

 

= (1.86 ℃/m) (0.752 m)

= 1.399 ℃

≒ 1.40 ℃

 

 

 

물의 정상 어는점 = 0℃ 이므로,

답: B) –1.41℃

 

 

 

 

[키워드] fructose 1.049 g/mL osmotic 17.0 atm 25℃ freeze