sucrose와 NaCl 혼합물의 질량백분율과 몰분율. 어는점내림

sucrose와 NaCl 혼합물의 질량백분율과 몰분율. 어는점내림

 

 

You take 20.0 g of a sucrose (C12H22O11) and NaCl mixture and dissolve it in 1.0 L of water. The freezing point of this solution is found to be -0.426°C. Assuming ideal behavior, calculate the mass percent composition of the original mixture, and the mole fraction of sucrose in the original mixture.

 

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물의 Kf = 1.86 ℃/m

 

ΔTf = Kf • m

0.426 ℃ = (1.86 ℃/m)(?)

? = (0.426 ℃) / (1.86 ℃/m)

= 0.229 m = 0.229 mol/kg

 

(0.229 mol/kg) × 1 kg = 0.229 mol ... 용액에 용해되어있는 전체 혼합물의 몰수

 

NaCl의 질량을 x라 두면, sucrose의 질량 = 20.0-x

 

[(20.0-x) g / (342.30 g/mol)] + [2 × (x g / (58.44 g/mol))] = 0.229 mol

( 단, NaCl(aq) → Na^+(aq) + Cl^-(aq) 이므로, van't Hoff 인자(i) = 2 )

 

20.0/342.30 – x/342.30 + 2(x/58.44) = 0.229

0.0584 – 0.00292x + 0.0342x = 0.229

0.0313x = 0.229 – 0.0584 = 0.171

x = 0.171 / 0.0313 = 5.46 g NaCl

20.0 – 5.46 = 14.54 g sucrose

 

sucrose의 질량 백분율 = (14.54 / 20.0) × 100 = 72.7%

NaCl의 질량 백분율 = (5.46 / 20.0) × 100 = 27.3%

 

sucrose의 몰수 = 14.54 g / (342.30 g/mol) = 0.0425 mol

NaCl의 몰수 = 5.46 g / (58.44 g/mol) = 0.0934 mol

 

sucrose의 몰분율 = 0.0425 / ( 0.0934 + 0.0425) = 0.313

NaCl의 몰분율 = 0.0934 / ( 0.0934 + 0.0425) = 0.687

 

 

답: 72.7% sucrose, 27.3% NaCl, 0.313