Question

What is the mole fraction of LiF in an aqueous solution that contains 39.4% LiF?(w/w%)

Answer 1

`chi_(LiF)=0.311`

Explanation:

We work from a `100*g` mass of the SOLUTION...and interrogate its molar composition...

`chi_(LiF)="Moles of salt"/"Moles of salt AND moles of water"`

`=((39.4*g)/(25.94*g*mol^-1))/((39.4*g)/(25.94*g*mol^-1)+(100.0*g-39.4*g)/(18.01*g*mol^-1))=(1.52*mol)/(1.52*mol+3.37*mol)`

`=0.311`

What is `chi_"water"`?