Question

What is the mole fraction, X, of solute and the molarity m, for an aqueous solution that is 16.0% NaOH by mass?

Answer 1

Well, `chi_"the mole fraction"="moles of solute"/"total moles IN THE solution"`

Explanation:

And so we take a `100*g` mass of solution.....of which `16.0*g` is sodium hydroxide, and `84.0*g` is water.

And thus `chi_"NaOH"=((16.0*g)/(40.0*g*mol^-1))/(((16.0*g)/(40.0*g*mol^-1)+(84.0*g)/(18.01*g*mol^-1))`

`=0.0790`...

And `chi_"water"=((84.0*g)/(18.01*g*mol^-1))/(((16.0*g)/(40.0*g*mol^-1)+(84.0*g)/(18.01*g*mol^-1)))=0.921`

And `chi_"water"+chi_"NaOH"=0.921+0.0790=1` as is absolutely required for a binary solution. Capisce?

And as for `"molarity"`, WE ASSUME that the solution has a volume of `84.0*mL`...and so we take the quotient...

`"molarity"="moles of solute"/"volume of solution"`.

`=((16.0*g)/(40.0*g*mol^-1))/(84.0*gxx1*mL*g^-1xx10^-3*L*mL^-1)=4.76*mol*L^-1`...