Description
An aqueous antifreeze solution is 31.0% ethylene glycol (C2H6O2) by mass.
The density of the solution is 1.039 g/cm3.
Calculate the mole fraction of ethylene glycol.
Explanation & Answer
C2H6O10
Atomic mass of C =12
Atomic mass of H =1
Atomic mass of O =16
Atomic mass of C2H6O10 = 2*12+6*1+16*2 =62
C2 H6 O10 =0.31*1.039= 0.322
Water = 1.039-0.322 =0.717
Moles of C2 H6 O10 =0.322/62 =0.00519
Moles of water = 0.717/18 =0.0398
Mole fraction of Glycol = 0.00519/(0.00519+0.0398) = 0.115
Mole fraction of water = 0.0398/(0.00519+0.0398) = 0.885
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