Question

What is the boiling point of a 0.321m aqueous solution of `NaCl`?

Answer 1

We need the `"molal boiling point elevation constant"` for water.........and thus the boiling point of the solution is `100.33` `""^@C`.

Explanation:

This site gives `K_B=0.512*K*kg*mol^-1` as the so-called `"ebullioscopic constant"` for water. Of course there is a catch, and this requires us to consider ALL the species in solution, i.e. in the given solution we have concentrations of `0.321*molal` with respect to BOTH `Na^+` and `Cl^-` ions.......

Now `DeltaT_b=K_BxxCxxi`, where `i` is the number of ions in solution......

`=0.512*K*kg*mol^-1xx0.321*mol*kg^-1xx2`

`=0.33*K`; and this is a boiling point elevation with respect to the PURE solvent......And thus the boiling point of the solution is `100.33` `""^@C`.