Question

Question #2177b

Answer 1

`chi_(NaOH)~=0.05`

Explanation:

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

And our `NaOH(aq)` solution is `3.0*"molal"`, i.e. `3.0*mol*kg^-1`

Now `chi_(NaOH)="Moles of NaOH"/"Moles of NaOH + moles of water"`

`=(3.0*mol)/(3.0*mol+(1000*g)/(18.01*g*mol^-1))`

`=(3.0*mol)/(3.0*mol+55.52*mol)=0.0513`