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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.

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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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