Using the data... BHT only freezing point: 68.36 Celsius Freezing Point BHT & Ce
ID: 676779 • Letter: U
Question
Using the data... BHT only freezing point: 68.36 Celsius Freezing Point BHT & CetylAlcohol: 60.59 Celsius Freezing Depression BHT & Cetyl Alcohol: 4.8 CelsiusFreezing Point BHT & Unknown: 61.52 Celsius Freezing Depression BHT & Unknown: 6.84 Celsius Weight BHT: 6.073g Weight Cetyl Alcohol: 0.755g Weight Unknown: 1.045g c. Calculate the Kf using the data from the cetyl alcohol trials. Then use the Kf todetermine the molar mass of the unknown. (Kg of solvent =0.006073kg BHT). Using the data... BHT only freezing point: 68.36 Celsius Freezing Point BHT & CetylAlcohol: 60.59 Celsius Freezing Depression BHT & Cetyl Alcohol: 4.8 Celsius
Freezing Point BHT & Unknown: 61.52 Celsius Freezing Depression BHT & Unknown: 6.84 Celsius Weight BHT: 6.073g Weight Cetyl Alcohol: 0.755g Weight Unknown: 1.045g c. Calculate the Kf using the data from the cetyl alcohol trials. Then use the Kf todetermine the molar mass of the unknown. (Kg of solvent =0.006073kg BHT).
Explanation / Answer
first you know that =Kf*m now plug 68.36-60.59=4.8*m solve for m then u know thats moles solute / Kg solvent. mutliply by kg of bht .006073 and you get moles of solute cetylalchol. then divide grams of cetyl achohol .755/ moles taht you just solvedfor to get molar mass of cetyl achohol.
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