10.0 g liquid 90.0 g water fp –3.33℃ mol mass

10.0 g liquid 90.0 g water fp –3.33℃ mol mass

 

 

A solution containing 10.0 g of an unknown liquid

and 90.0 g water has a freezing point of –3.33℃.

Given Kf = 1.86 ℃/m for water,

the molar mass of the unknown liquid is _____ g/mol.

 

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ΔTf = Kf × m

( 참고 https://ywpop.tistory.com/1920 )

 

 

 

순수한 물의 어는점 = 0℃ 이므로,

ΔTf = 0 – (–3.33) = 3.33℃

 

 

 

m = ΔTf / Kf

= (3.33℃) / (1.86 ℃/m)

= 1.79 mol/kg

 

 

 

1.79 mol/kg = ? mol / (90.0/1000 kg)

 

? = 1.79 × (90.0/1000) = 0.161 mol

---> 미지 액체 용질의 몰수

 

 

 

몰질량을 계산하면,

10.0 g / 0.161 mol = 62.1 g/mol

( 참고 https://ywpop.tistory.com/7738 )

 

 

 

답: 62.1