A mixture has 18 g water and 414 g ethanol. What are the mole fractions of water and ethanol?

Given : 

Mass of water = 18 g, 

Mass of ethanol = 414 g

To find : 

Mole fractions of water and ethanol

Formulae :

Mole fractions of A = \(\frac{n_A}{n_A+n_B}\) 

Mole fractions of B = \(\frac{n_B}{n_A+n_B}\)  

Calculation :

Number of moles of water (n_A) = \(\frac{18}{18}\) = 1 mol

Number of moles of ethanol (n_B) = \(\frac{414}{46}\) = 9 mol

Mole fraction of water (A) = \(\frac{n_A}{n_A+n_B}\)= \(\frac{1}{1+9}\) 

= \(\frac{1}{10}\) 

= 0.1

Mole fraction of ethanol (B) = \(\frac{n_B}{n_A+n_B}\)= \(\frac{9}{1+9}\) 

= \(\frac{9}{10}\) 

= 0.9

∴ Mole fraction of water = 0.1

Mole fraction of ethanol = 0.9