Answer
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%
