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 the freezing point  TF for T- BUTANOL  is 25.50 and Kf is 9.1 c/m . usually...

 the freezing point  TF for T- BUTANOL  is 25.50 and Kf is 9.1 c/m . usually t butanol absorbs water  on exposure to the air. if the freezzing  point of a 10.0 g sample t butanol is measured as 24.59 degree celcius  , how many grams of water are present in the sample
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Answer #1
ΔT = 25.50 - 24.59 = 0.91 C

ΔT = Kf*m
m = ΔT/Kf = 0.91 C /(9.1 C/m) = 0.1 molal = 0.1 molwater/ kg butanol

10 g butanol *(0.1 mol water / kg butanol) *(1 kg/1000g) *(18g/1mol water)
= 0.018 grams water
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