Time remaining:
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

Are you studying on the go? Check out our FREE app and post questions on the fly!
Download on the
App Store
...
Mar 13th, 2015
...
Mar 13th, 2015
Dec 3rd, 2016
check_circle
Mark as Final Answer
check_circle
Unmark as Final Answer
check_circle
Final Answer

Secure Information

Content will be erased after question is completed.

check_circle
Final Answer