Chemistry and Chemical Reactivity (9th Edition)

Published by Cengage Learning
ISBN 10: 1133949649
ISBN 13: 978-1-13394-964-0

Chapter 13 Solutions and Their Behavior - Study Questions - Page 505: 2


data given: mass of camphor in g = 45 g volume of ethanol =425 ml density of ethanol =0.785 g\ml molality of camphor= ? mole fraction of camphor= ? weight percentage of camphor = ? solution: firstly find moles of camphor and ethanol : moles = mass in g /molar mass molar mass of camphor C10H160 = 12*10+1*16+16 =152 molar mass of ethanol C2H5OH = 12*2+1*5+16+1 =46 moles of camphor = 45/152 =0.296g/mol in order to find moles of ethanol we need mass of ethanol in g, so we use density density=mass/volume mass= density *volume =0.785*425= 333.625 g moles of ethanol= 333.625/46 =7.252 g/mol now molality = moles of solute(camphor) /liters of solvent(ethanol) we have ethanol in ml as 1l=1000ml 425ml= 0.425l molality = 0.296/ 0.425 = 0.696 m now mole fraction= moles of camphor/( moles of camphor +moles of ethanol) =0.296/(0.296+7.252) =0.03921 weight percentage = mass of camphor/ (mass of camphor+mass of ethanol ) *100 =45/(45+333.625)*100 =11.885%
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