Question

What are the mole fractions of CH3OH and H2O in the solution that contains 128g of CH3OH and 108g of H2O?

Answer 1

`chi_"MeOH"=0.40....`

Explanation:

`chi_"MeOH"=((128*g)/(32.04*g*mol^-1))/((128*g)/(32.04*g*mol^-1)+(108*g)/(18.01*g*mol^-1))=0.400`

Now we know that in a binary solution, the sum of the mole fractions will be one....but let us belabour the point....

`chi_"water"=((108*g)/(18.01*g*mol^-1))/((128*g)/(32.04*g*mol^-1)+(108*g)/(18.01*g*mol^-1))=0.600`

And `chi_"MeOH"+chi_"water"=1` as required....