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Title: The freezing point (How many grams.of water are present in the sample?) Post by: nevada on Dec 15, 2013 The freezing point of t-butanol is 25.5C and Kf is 9.lC.kg/mol. Usually t-butanol absorbs water on exposure to air. If the freezing point of a 10g sample of t-butanol is 24.5C. How many grams.of water are present in the sample?
i am trying to understand the steps to this question but the way my teacher did it, is confusing me. ∆T=Kf*m ∆T=25.5-24.59=0.91 0.91=9.1(m) m=0.1mol H2O mol H2O=massH2O/ MWH2O= x/ 18g kg solvent= (10-x) * 1kg/1000g 0.1=x/18 / (10-x)/1000 =1000X / 180-18x 18-1.8X=1000X x=0.018g H2O can you please explain how to do this? thank you! Title: Re: The freezing point (How many grams.of water are present in the sample?) Post by: padre on Dec 15, 2013 Content hidden
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