Chemistry Help How to get the mole fraction ?

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To calculate the mole fraction of ethylene glycol in a 37.0% aqueous antifreeze solution with a density of 1.047 g/cm3, the correct approach involves determining the number of moles of both ethylene glycol and water. The molar mass of ethylene glycol is 62 g/mol, and that of water is 18 g/mol. The mole fraction formula is X = nA/(nA+nB), where nA is the moles of ethylene glycol and nB is the moles of water. The initial calculations led to incorrect percentages, indicating a misunderstanding of how to express the result, as the system requires a mole fraction rather than a percentage. Accurate calculations should yield a mole fraction value, which can be derived from the correct mass and density inputs.
zhen
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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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