Calculation of Solute Mole Fraction

***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?***




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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.)




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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