If #4.57*g# of glucose are dissolved in #25.2*g# of water, what are the mole fractions, #chi_"each component"#?

1 Answer
Jun 1, 2017

Answer:

#"Mole fraction,"# #chi="Moles of solute"/"Total moles of solution"#

Explanation:

And thus the sum of the mole fractions must be equal to one.......

#chi_"glucose"=((4.75*g)/(180.16*g*mol^-1))/((4.75*g)/(180.16*g*mol^-1)+(25.2*g)/(18.01*g*mol^-1))=0.0185#

#chi_"water"=((25.2*g)/(18.01*g*mol^-1))/((4.75*g)/(180.16*g*mol^-1)+(25.2*g)/(18.01*g*mol^-1))=0.9815#

And these are dimensionless quantities............