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Suppose we had an aqueous solution of density 0.984 g cm$^{-3}$, containing 100.0 g ethanol per liter. How would we compute the mole fraction of ethanol in this solution?


My calculations so far propose that we examine a fixed volume, 1 L of the solution:

CH$_3$CH$_2$OH: m = 100.0g | M = 46.068g mol$^{-1}$ | n = $\frac{m}{M}$ = 2.170 ... mol

solution (incl. H$_2$O): m$_{tot}$ = 984 g | M$_{tot}$ = M$_{solute}$ + M$_{solvent}$ = 46.068 + 18.016 = 64.084 g mol$^{-1}$

$\therefore$ n$_{tot}$ = 15.35 ... mol

Hence $x$ = $\frac{n}{n_{tot}}$ = 0.1413 ... $\approx$ 0.141 (3 sig. fig.)

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This is not correct because you assuming that the molecular weight rations are one to one.

The coorect formulation is

(100 g Methano/46.068)/(100/46.068 + (984-100)/18.016)

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