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Calculating mole fraction, mass percent, molality

Chemistry
Tutor: None Selected Time limit: 0 Hours

Mar 13th, 2015

a) Molar mass of Phenol (C6H5OH) = 12.0107*6+1.00794*5+15.9994+1.00794

  = 94.11124 g/mol

& molar mass of Ethanol (CH3CH2OH) = 12.0107+ 1.00794*3+ 12.0107*1.00794*2 + 15.9994 + 1.00794

  = 46.06844 g/mol

So, Phenol (C6H5OH) = 28.5/94.11124 = 0.30283 mol & Ethanol (CH3CH2OH) = 516/46.06844 = 11.20072 mol

Total = 0.30283+11.20072 = 11.50355 mol

So, mole fraction of Phenol (C6H5OH) = 0.30283/11.50355= 0.02632 (Answer)

b)  Mass percent of Phenol (C6H5OH) =28.5 X 100/(28.5+516) = 5.23% (Answer)

c)  As, Molality = moles of solute / kilograms of solvent

So, molality of Phenol (C6H5OH)= 0.30283/ (516*1000) = 0.587 m (Answer)


Mar 14th, 2015

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