Help How to get the mole fraction ?

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SUMMARY

The discussion centers on calculating the mole fraction of ethylene glycol (C2H6O2) in a 37.0% aqueous antifreeze solution with a density of 1.047 g/cm³. The correct formula for mole fraction is X = nA/(nA+nB), where nA is the number of moles of ethylene glycol and nB is the number of moles of water (H2O). The user initially calculated a mole fraction of 13.8% and later 14.6%, both of which were incorrect due to misunderstanding the output format required by the system, which does not accept percentages.

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An aqueous antifreeze solution is 37.0% ethylene glycol (C2H6O2) by mass. The density of the solution is 1.047 g/cm3.

Q: Calculate the mole fraction of ethylene glycol.

I tried it many time, but still got the wrong answer.

that is what i did:

mole fraction:(X) = nA/(nA+nB)
set the mass of the solution - m
molar mass of C2H6O2: 62
molar mass of H2O: 18 --- (I think the solvent should be H2O...but not quite sure)

=> [(0.37 * m)/62]/[(0.37 * m/62 + 0.63 * m/18)]
= 13.8%------but that is wrong

need help...please
 
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Check your math - equation is OK, just the result is wrong.
 
oh...never mind...
the system does not accept percentage...
 

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