Question

What is the mole fraction of solute in a 3.89 m aqueous solution?

Answer 1

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

`chi=(3.89*mol)/(3.89*mol+(1000*g)/(18.01*g*mol^-1))=0.067`

Explanation:

And `chi,` `"mole fraction"` `=` `"Moles of solute"/"Moles of solute and moles of solvent"`

We have a `3.89` `"molal"` solution (at least I think we do); i.e. `"molality"` `=` `3.89*mol*kg^-1`.

And thus `chi=(3.89*mol)/(3.89*mol+(1000*g)/(18.01*g*mol^-1))=0.067`