What are the mole fractions of each component prepared by adding #5*g# of water to #95*g# of ethanol?

Question

Ellie Andrews · Accepted Answer

#chi_"component"="moles of component"/"total moles of mixture"....#

We work from a #100*g# mass of solution....

And so #chi_"EtOH"=((95*g)/(46.07*g*mol^-1))/((95*g)/(46.07*g*mol^-1)+(5*g)/(18.01*g*mol^-1))=0.881#

And likewise #chi_"water"==((5*g)/(18.01*g*mol^-1))/((95*g)/(46.07*g*mol^-1)+(5*g)/(18.01*g*mol^-1))=??#

In a binary solution, should the mole fractions add up to ONE?

Cameron Andrews · Answer

undefined To find the mole fractions of each component, we first need to calculate the number of moles of each substance.

1. Calculate the number of moles of water:
   $$ \text{moles of water} = \frac{\text{mass of water}}{\text{molar mass of water}} $$

2. Calculate the number of moles of ethanol:
   $$ \text{moles of ethanol} = \frac{\text{mass of ethanol}}{\text{molar mass of ethanol}} $$

3. Calculate the total moles:
   $$ \text{total moles} = \text{moles of water} + \text{moles of ethanol} $$

4. Calculate the mole fraction of each component:
   $$ \text{Mole fraction of water} = \frac{\text{moles of water}}{\text{total moles}} $$
   $$ \text{Mole fraction of ethanol} = \frac{\text{moles of ethanol}}{\text{total moles}} $$

Substitute the given masses of water and ethanol, along with their respective molar masses, into the above equations to find the mole fractions.