Question

What are the molal concentrations, and mole fractions, of a `385*mg` mass of `C_26H_46O` dissolved in `40*mL` of chloroform?

Answer 1

`"Molality"` `=` `0.0249*mol*kg^-1`

Explanation:

`"Molality"` `=` `"Moles of solute"/"Kilograms of solvent"`

And here, `"molality"` `=` `((0.385*g)/(386.67*g*mol^-1))/(0.040*kg)=0.0249*mol*kg^-1`.

On the other hand, the mole fraction, `chi`, is dimensionless, and this is defined by the quotient.........

`chi_"solute"="Moles of solute"/"Total number of moles in solution"`

So here.........

`chi_"cholesterol"=("Moles of cholesterol")/("Moles of cholesterol+moles of chloroform")`

`=((0.385*g)/(386.67*g*mol^-1))/((0.385*g)/(386.67*g*mol^-1)+(40*g)/(119.38))=2.96xx10^-3`

And `"NECESSARILY"` `chi_"chloroform"=1-chi_"cholesterol"=0.997`

Why `"necessarily?"`